Given, E°_{Cr^{3+}/Cr} = -0.72 V, E°_{Fe^{2+}/Fe} = -0.42 V. The potential for the cell Cr | Cr^{3+}

Q: Why do we multiply the half-reactions by 2 and 3?

In any redox reaction, the number of electrons lost in oxidation must exactly equal the number of electrons gained in reduction. Chromium loses 3 electrons, while Iron gains 2. The lowest common multiple is 6. So, we multiply the Chromium reaction by 2 and the Iron reaction by 3 to balance the electron transfer.

Q: Does multiplying the half-reaction change the standard electrode potential?

No! Standard electrode potential ($E^{\circ}$) is an intensive property, meaning it does not depend on the amount of substance. Even if you multiply the reaction by 100, the $E^{\circ}$ value remains exactly the same.

Q: How do we know which electrode is the anode and which is the cathode?

In standard cell notation (like $A | A^+ || B^+ | B$), the anode (where oxidation occurs) is always written on the left side of the double vertical lines (the salt bridge), and the cathode (where reduction occurs) is always written on the right.

Q: Why is the concentration of solid electrodes not included in the Nernst equation?

The concentration of a pure solid or a pure liquid is constant and is conventionally taken as unity (1). Therefore, they do not affect the reaction quotient $Q$ and are omitted from the Nernst equation.

Q: What happens if the cell potential comes out to be negative?

A positive cell potential ($E_{cell} > 0$) indicates that the cell reaction is spontaneous as written, and the cell can act as a galvanic cell to produce electricity. If it comes out negative, it means the reaction is non-spontaneous in the forward direction, and you would need to supply an external voltage greater than that magnitude to force the reaction to occur (electrolytic cell).

Question

Given, $E^{\circ}_{Cr^{3+}/Cr} = -0.72	ext{ V}$, $E^{\circ}_{Fe^{2+}/Fe} = -0.42	ext{ V}$.
The potential for the cell $Cr | Cr^{3+} (0.1	ext{ M}) || Fe^{2+} (0.01	ext{ M}) | Fe$ is

Options:

(A) 0.26 V

(B) 0.399 V

(C) - 0.339 V

(D) - 0.26 V

Sigma Percentile · Accepted Answer

### The Electrochemical Setup

Imagine you are standing in a laboratory, looking at a classic electrochemical cell. On the left, you have a solid Chromium electrode submerged in a solution of $Cr^{3+}$ ions with a concentration of $0.1	ext{ M}$. On the right, there is an Iron electrode resting in a solution of $Fe^{2+}$ ions at a concentration of $0.01	ext{ M}$.

The standard cell notation $Cr | Cr^{3+} (0.1	ext{ M}) || Fe^{2+} (0.01	ext{ M}) | Fe$ tells us a story. The double vertical lines represent the salt bridge, dividing the cell into two halves. By convention, the anode (where oxidation occurs) is written on the left, and the cathode (where reduction occurs) is on the right.

### Balancing the Electrons

Before we can calculate anything, we need to understand the chemistry happening at each electrode.

At the anode, Chromium metal is losing electrons to become $Cr^{3+}$ ions:
$$Cr ightarrow Cr^{3+} + 3e^{-}$$

At the cathode, $Fe^{2+}$ ions are gaining electrons to form solid Iron:
$$Fe^{2+} + 2e^{-} ightarrow Fe$$

Notice a problem? The anode produces 3 electrons, but the cathode only consumes 2. Nature demands balance. To equalize the electron transfer, we must multiply the oxidation reaction by 2 and the reduction reaction by 3.

This gives us the balanced net cell reaction:
$$2Cr + 3Fe^{2+} ightarrow 2Cr^{3+} + 3Fe$$

Crucially, this tells us that the total number of electrons transferred in the balanced reaction, $n$, is $6$.

### The Standard Cell Potential

Next, we determine the standard cell potential, $E^{\circ}_{cell}$. This is the potential the cell would have if all concentrations were exactly $1	ext{ M}$.

The formula is simple:
$$E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode}$$

We are given the standard reduction potentials: $E^{\circ}_{Fe^{2+}/Fe} = -0.42	ext{ V}$ and $E^{\circ}_{Cr^{3+}/Cr} = -0.72	ext{ V}$.

Substituting these values:
$$E^{\circ}_{cell} = -0.42 - (-0.72) = 0.30	ext{ V}$$

### The Master Equation: Nernst

Because our concentrations are not $1	ext{ M}$, the standard potential isn't the whole story. We must use the **Nernst Equation** to find the actual cell potential, $E_{cell}$:
$$E_{cell} = E^{\circ}_{cell} - \frac{0.0591}{n} \log Q$$

Here, $Q$ is the reaction quotient. It is the ratio of the concentrations of the products to the reactants, raised to the power of their stoichiometric coefficients. Remember, pure solids like $Cr$ and $Fe$ are excluded from $Q$.
$$Q = \frac{[Cr^{3+}]^2}{[Fe^{2+}]^3}$$

### The Final Calculation

Now, we substitute our known values into the Nernst equation.

$$E_{cell} = 0.30 - \frac{0.0591}{6} \log \frac{(0.1)^2}{(0.01)^3}$$

Let's simplify the logarithmic term carefully. 
$$(0.1)^2 = (10^{-1})^2 = 10^{-2}$$
$$(0.01)^3 = (10^{-2})^3 = 10^{-6}$$

So, the fraction becomes:
$$\frac{10^{-2}}{10^{-6}} = 10^4$$

Substituting this back into our equation:
$$E_{cell} = 0.30 - \frac{0.0591}{6} \log(10^4)$$

Since $\log(10^4) = 4$, we have:
$$E_{cell} = 0.30 - \frac{0.0591 	imes 4}{6}$$
$$E_{cell} = 0.30 - 0.0394 = 0.2606	ext{ V}$$

Rounding to two decimal places, we get **$0.26	ext{ V}$**, which perfectly matches option (a).

Animated Solution for Chemistry - Electrochemistry: 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

$C e l l N o t a t i o n$

$H a l f - C e l l R e a c t i o n s$

$S t an d a r d C e l l P o t e n t ia l$

$N er n s t E q u a t i o n$

$S u b s t i t u t i o n$

$S im pl i f i c a t i o n$

$F ina l C a l c u l a t i o n$

The Sigma Insight: Electrochemical Cells

The Electrochemical Setup

Balancing the Electrons

The Standard Cell Potential

The Master Equation

The Final Calculation

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Given below are the half-cell reactions $M n^{2 +} + 2 e^{-} ⟶ M n, E^{\circ} = - 1.18 V$ $2 (M n^{3 +} + e^{-} ⟶ M n^{2 +}), E^{\circ} = + 1.51 V$ The $E^{\circ}$ for $3 M n^{2 +} ⟶ M n + 2 M n^{3 +}$ will be

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Visualized Solution

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Half−Cell Reactions

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

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

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

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

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