Some standard electrode potentials at 298 K are given below:
Pb2+/Pb -0.13 V
Ni2+/Ni -0.24 V
Cd2+/

Q: Why is rod X the anode?

Dissolution means the solid metal is turning into aqueous ions by losing electrons. Loss of electrons is oxidation, which always occurs at the anode.

Q: Why did we use 0.06 instead of 0.0591 in the Nernst equation?

In JEE problems, unless extreme precision is required, 0.0591 is often approximated to 0.06 to make manual calculations faster and less error-prone.

Q: Can a cell work if its standard potential $E^\circ_{\text{cell}}$ is negative?

Yes! As seen in option (B), a negative standard potential can be overcome if the concentration gradient (the $\log Q$ term) provides enough positive driving force to make the overall $E_{\text{cell}}$ positive.

Q: What happens to the cell potential as the reaction proceeds?

As X dissolves, $[X^{2+}]$ increases, and as Y deposits, $[Y^{2+}]$ decreases. This increases $Q$, making the $\log Q$ term more positive, which decreases $E_{\text{cell}}$ until it reaches zero (equilibrium).

Q: Why don't we include the solid metals in the reaction quotient $Q$?

The activity (or effective concentration) of pure solids is constant and taken as 1, so they do not appear in the equilibrium expression or the reaction quotient.

Question

Some standard electrode potentials at 298 K are given below:
$	ext{Pb}^{2+}/	ext{Pb} \quad -0.13 	ext{ V}$
$	ext{Ni}^{2+}/	ext{Ni} \quad -0.24 	ext{ V}$
$	ext{Cd}^{2+}/	ext{Cd} \quad -0.40 	ext{ V}$
$	ext{Fe}^{2+}/	ext{Fe} \quad -0.44 	ext{ V}$
To a solution containing $0.001 	ext{ M}$ of $	ext{X}^{2+}$ and $0.1 	ext{ M}$ of $	ext{Y}^{2+}$, the metal rods X and Y are inserted (at 298 K) and connected by a conducting wire. This resulted in dissolution of X. The correct combination(s) of X and Y, respectively, is (are)
(Given: Gas constant, $	ext{R} = 8.314 	ext{ J K}^{-1} 	ext{ mol}^{-1}$, Faraday constant, $	ext{F} = 96500 	ext{ C mol}^{-1}$)

Options:

(A) Cd and Ni

(B) Cd and Fe

(C) Ni and Pb

(D) Ni and Fe

Sigma Percentile · Accepted Answer

### The Battle of the Metals: Unraveling Electrochemical Spontaneity

Imagine you are standing in a laboratory, observing a fascinating duel between two metals. You have a single beaker containing a mixture of two different metal ions: $X^{2+}$ at a very dilute concentration of $0.001 	ext{ M}$, and $Y^{2+}$ at a much higher concentration of $0.1 	ext{ M}$. You plunge two solid metal rods, X and Y, into this chemical arena and connect them with a conducting wire.

Suddenly, you notice something crucial: **Rod X begins to dissolve**.

This simple physical observation is the key that unlocks the entire problem. In electrochemistry, when a solid metal dissolves into a solution, it is losing electrons and transforming into aqueous ions. This process is called **oxidation**. Because oxidation always occurs at the anode, we can definitively state that Rod X is acting as our anode, and by elimination, Rod Y is our cathode where reduction takes place.

### The Master Equation: Nernst in Action

To determine which metals could possibly play the roles of X and Y, we need to calculate the overall cell potential, $E_{	ext{cell}}$. For a reaction to occur spontaneously (like the dissolution of Rod X), the overall cell potential must be strictly positive ($E_{	ext{cell}} > 0$).

Because our ionic concentrations are not at the standard $1 	ext{ M}$, we cannot simply rely on standard electrode potentials ($E^\circ_{	ext{cell}}$). We must invoke the mighty **Nernst Equation**, which bridges the gap between standard conditions and the reality of our beaker:

$$E_{	ext{cell}} = E^\circ_{	ext{cell}} - \frac{0.0591}{n} \log Q$$

Here, $n$ is the number of electrons transferred. Since both metals form divalent ions ($X^{2+}$ and $Y^{2+}$), $n = 2$. The reaction quotient, $Q$, is the ratio of the concentration of the products to the reactants. Based on our identified anode and cathode, the overall cell reaction is:

$$X(s) + Y^{2+}(aq) ightarrow X^{2+}(aq) + Y(s)$$

Therefore, $Q = \frac{[X^{2+}]}{[Y^{2+}]}$.

### Crunching the Numbers

Let's substitute our known values into the Nernst equation. To make calculations smoother, it is a standard practice in JEE problems to approximate $0.0591$ to $0.06$:

$$E_{	ext{cell}} = E^\circ_{	ext{cell}} - \frac{0.06}{2} \log \left( \frac{0.001}{0.1} ight)$$

Simplifying the fraction inside the logarithm gives us $10^{-2}$. The base-10 logarithm of $10^{-2}$ is simply $-2$.

$$E_{	ext{cell}} = E^\circ_{	ext{cell}} - 0.03 	imes (-2)$$

$$E_{	ext{cell}} = E^\circ_{	ext{cell}} + 0.06 	ext{ V}$$

This is a profound result! The concentration gradient is actually *helping* the reaction move forward, providing a $+0.06 	ext{ V}$ boost to the standard cell potential. For the dissolution to be spontaneous, we need $E_{	ext{cell}} > 0$, which means:

$$E^\circ_{	ext{cell}} + 0.06 > 0 \implies E^\circ_{	ext{cell}} > -0.06 	ext{ V}$$

### Testing the Contenders

Now, we simply test each option by calculating its standard cell potential, $E^\circ_{	ext{cell}} = E^\circ_{	ext{cathode}} - E^\circ_{	ext{anode}} = E^\circ_{Y} - E^\circ_{X}$.

**Option (A): X = Cd, Y = Ni**
$$E^\circ_{	ext{cell}} = -0.24 - (-0.40) = +0.16 	ext{ V}$$
$$E_{	ext{cell}} = 0.16 + 0.06 = +0.22 	ext{ V}$$
Since $0.22 	ext{ V} > 0$, this combination works perfectly.

**Option (B): X = Cd, Y = Fe**
$$E^\circ_{	ext{cell}} = -0.44 - (-0.40) = -0.04 	ext{ V}$$
$$E_{	ext{cell}} = -0.04 + 0.06 = +0.02 	ext{ V}$$
Look closely here! Even though the standard potential is negative, the concentration boost pushes the overall potential into the positive territory. This combination is also correct.

**Option (C): X = Ni, Y = Pb**
$$E^\circ_{	ext{cell}} = -0.13 - (-0.24) = +0.11 	ext{ V}$$
$$E_{	ext{cell}} = 0.11 + 0.06 = +0.17 	ext{ V}$$
Since $0.17 	ext{ V} > 0$, this combination is a winner as well.

**Option (D): X = Ni, Y = Fe**
$$E^\circ_{	ext{cell}} = -0.44 - (-0.24) = -0.20 	ext{ V}$$
$$E_{	ext{cell}} = -0.20 + 0.06 = -0.14 	ext{ V}$$
Here, the standard potential is too negative. The $+0.06 	ext{ V}$ boost isn't enough to overcome it. The overall potential remains negative, meaning Rod X will not dissolve spontaneously. This option is incorrect.

### Conclusion: The Power of Concentration

This problem beautifully illustrates that chemical spontaneity isn't just about the inherent nature of the metals (their standard potentials). The environment—specifically, the concentration of ions—plays a massive role. By keeping the product concentration low and the reactant concentration high, we can force a seemingly non-spontaneous reaction to occur!

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Visualized Solution

$V i s u a l i z in g t h e E l ec t r oc h e mi c a l C e l l$

$I d e n t i f y in g A n o d e an d C a t h o d e$

$S e tt in g u p t h e N er n s t E q u a t i o n$

$S im pl i f y in g t h e L o g a r i t hmi c T er m$

$C o n d i t i o n f or S p o n t an e i t y$

$E v a l u a t in g O pt i o n (A)$

$E v a l u a t in g O pt i o n (B)$

$E v a l u a t in g O pt i o n (C)$

$E v a l u a t in g O pt i o n (D)$

$F ina l C o n c l u s i o n$

$T h e W a y F or w a r d$

The Sigma Insight: Electrochemical Cells

The Battle of the Metals

The Master Equation

Crunching the Numbers

Testing the Contenders

Conclusion

Similar Questions

The standard reduction potentials for $Zn^{2 +} / Zn$ , $Ni^{2 +} / Ni$ and $Fe^{2 +} / Fe$ are $- 0.76$ , $- 0.23$ and $- 0.44 V$ , respectively. The reaction $X + Y^{2 +} ⟶ X^{2 +} + Y$ will be spontaneous when

Given, $E_{F e^{3 +} / F e}^{\circ} = - 0.036 V$ , $E_{F e^{2 +} / F e}^{\circ} = - 0.439 V$ . The value of standard electrode potential for the charge, $F e^{3 +} (a q) + e^{-} ⟶ F e^{2 +} (a q)$ will be

Consider the following $E^{\circ}$ values :\n $E_{F e^{3 +} / F e^{2 +}}^{\circ} = + 0.77 V$ \n $E_{S n^{2 +} / S n}^{\circ} = - 0.14 V$ \nUnder standard conditions, the potential for the reaction\n $S n (s) + 2 F e^{3 +} (a q) \to 2 F e^{2 +} (a q) + S n^{2 +} (a q)$ is

If the standard electrode potential for a cell is $2 V$ at $300 K$ , the equilibrium constant ( $K$ ) for the reaction, $Zn (s) + Cu^{2 +} (a q) ⇌ Zn^{2 +} (a q) + Cu (s)$ at $300 K$ is approximately ( $R = 8 JK^{- 1} mol^{- 1}$ , $F = 96000 C mol^{- 1}$ )

The standard Gibbs energy for the given cell reaction in $kJ mol^{- 1}$ at $298 K$ is $Zn (s) + Cu^{2 +} (a q) ⟶ Zn^{2 +} (a q) + Cu (s), E^{\circ} = 2 V$ at $298 K$ (Faraday's constant, $F = 96000 C mol^{- 1}$ )

Given, $E_{C r^{3 +} / C r}^{\circ} = - 0.72 V$ , $E_{F e^{2 +} / F e}^{\circ} = - 0.42 V$ . The potential for the cell $C r ∣ C r^{3 +} (0.1 M) ∣∣ F e^{2 +} (0.01 M) ∣ F e$ is

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Visualizing the Electrochemical Cell

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Identifying Anode and Cathode

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Setting up the Nernst Equation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Simplifying the Logarithmic Term

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Condition for Spontaneity

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Evaluating Option (A)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Evaluating Option (B)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Evaluating Option (C)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Evaluating Option (D)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Conclusion

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Way Forward