For the cell, Zn(s)|Zn2+(aq)||Mx+(aq)|M(s), different half cells and their standard electrode potent

Q: Why do we divide the cell potential by $n$?

Because the question specifically asks for the maximum value of $E^{\circ}_{cell}$ *per electron transferred*. The variable $n$ represents the number of electrons transferred in the reduction half-reaction.

Q: Does multiplying a half-reaction by a coefficient change its $E^{\circ}$ value?

No, standard electrode potential ($E^{\circ}$) is an intensive property. It depends on the nature of the substance and its concentration, not on the amount of substance reacting.

Q: Why is $E^{\circ}_{anode}$ subtracted in the formula?

The formula $E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode}$ uses standard *reduction* potentials for both electrodes. Subtracting the anode's reduction potential is mathematically equivalent to adding its oxidation potential.

Q: What does a negative $E^{\circ}_{cell}$ mean?

A negative standard cell potential indicates that the overall redox reaction is non-spontaneous under standard conditions. In a working galvanic cell, $E^{\circ}_{cell}$ must be positive.

Q: How do we know the number of electrons transferred ($n$)?

You can determine $n$ by looking at the change in the oxidation state of the metal in the half-reaction. For example, in $Au^{3+} \rightarrow Au$, the oxidation state changes from +3 to 0, so $n = 3$.

Question

For the cell,
$Zn(s)|Zn^{2+}(aq)||M^{x+}(aq)|M(s)$,
different half cells and their standard electrode potentials are given below.

\begin{array}{|c|c|c|c|c|}
\hline
& Au^{3+}(aq)/Au(s) & Ag^{+}(aq)/Ag(s) & Fe^{3+}(aq)/Fe^{2+}(aq) & Fe^{2+}(aq)/Fe(s) \
\hline
E^{\circ}_{M^{x+}/M}/V & 1.40 & 0.80 & 0.77 & -0.44 \
\hline
\end{array}

If $E^{\circ}_{Zn^{2+}/Zn} = -0.76	ext{ V}$, which cathode will give a maximum value of $E^{\circ}_{cell}$ per electron transferred?

Options:

(A) $\frac{Ag^{+}}{Ag}$

(B) $\frac{Fe^{2+}}{Fe}$

(C) $\frac{Au^{3+}}{Au}$

(D) $\frac{Fe^{3+}}{Fe^{2+}}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are building a battery. On the left side, you have a Zinc electrode sitting in a solution of its own ions. This is our anode, where oxidation happens. On the right side, you have a mystery metal $M$ acting as the cathode, where reduction takes place.

The standard reduction potential of the Zinc anode is given as $E^{\circ}_{Zn^{2+}/Zn} = -0.76	ext{ V}$. Our mission is to choose the best cathode from a given list. But there is a twist! We are not just looking for the highest overall cell potential. We are looking for the maximum cell potential **per electron transferred**.

### The Master Equation

To find the standard cell potential, we use the fundamental equation of electrochemistry:

$$E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode}$$

Since $E^{\circ}_{anode} = -0.76	ext{ V}$, the equation becomes:

$$E^{\circ}_{cell} = E^{\circ}_{cathode} - (-0.76) = E^{\circ}_{cathode} + 0.76	ext{ V}$$

However, the question specifically asks us to maximize the ratio $\frac{E^{\circ}_{cell}}{n}$, where $n$ is the number of electrons transferred in the cathode half-reaction. Let's evaluate this ratio for each candidate.

### Evaluating the Candidates

**1. Gold ($Au^{3+}/Au$)**
The reduction potential is $1.40	ext{ V}$. The reaction is $Au^{3+} + 3e^{-} ightarrow Au$, so $n = 3$.

$$E^{\circ}_{cell} = 1.40 - (-0.76) = 2.16	ext{ V}$$

$$\frac{E^{\circ}_{cell}}{n} = \frac{2.16}{3} = 0.72	ext{ V}$$

**2. Silver ($Ag^{+}/Ag$)**
The reduction potential is $0.80	ext{ V}$. The reaction is $Ag^{+} + 1e^{-} ightarrow Ag$, so $n = 1$.

$$E^{\circ}_{cell} = 0.80 - (-0.76) = 1.56	ext{ V}$$

$$\frac{E^{\circ}_{cell}}{n} = \frac{1.56}{1} = 1.56	ext{ V}$$

**3. Iron(III) to Iron(II) ($Fe^{3+}/Fe^{2+}$)**
The reduction potential is $0.77	ext{ V}$. The reaction is $Fe^{3+} + 1e^{-} ightarrow Fe^{2+}$, so $n = 1$.

$$E^{\circ}_{cell} = 0.77 - (-0.76) = 1.53	ext{ V}$$

$$\frac{E^{\circ}_{cell}}{n} = \frac{1.53}{1} = 1.53	ext{ V}$$

**4. Iron(II) to Iron ($Fe^{2+}/Fe$)**
The reduction potential is $-0.44	ext{ V}$. The reaction is $Fe^{2+} + 2e^{-} ightarrow Fe$, so $n = 2$.

$$E^{\circ}_{cell} = -0.44 - (-0.76) = 0.32	ext{ V}$$

$$\frac{E^{\circ}_{cell}}{n} = \frac{0.32}{2} = 0.16	ext{ V}$$

### Final Conclusion

Comparing the calculated ratios, Silver ($Ag^{+}/Ag$) yields the highest value of **$1.56	ext{ V}$ per electron transferred**.

Notice how Gold had the highest *total* cell potential ($2.16	ext{ V}$), but because it requires three electrons per atom of Gold deposited, its potential *per electron* is much lower. This highlights the importance of reading the question carefully to avoid falling into a trap!

Select Answer:

Visualized Solution

$E_{ce l l}^{\circ} and n$

\text{Formula for } E^{\circ}_{cell}

\text{For } Au^{3+}/Au

\text{For } Ag^{+}/Ag

\text{For } Fe^{3+}/Fe^{2+}

\text{For } Fe^{2+}/Fe

\text{Conclusion}

\text{Food for Thought}

The Sigma Insight: Electrochemical Cells

Analyzing the Setup

The Master Equation

Evaluating the Candidates

Final Conclusion

Similar Questions

For the galvanic cell, $Z n (s) + C u^{2 +} (0.02 M) ⟶ Z n^{2 +} (0.04 M) + C u (s)$ , $E_{C e l l} = \dots \times 1 0^{- 2} V$ . (Nearest integer) [Use $E_{C u / C u^{2 +}}^{\circ} = - 0.34 V$ , $E_{Z n / Z n^{2 +}}^{\circ} = + 0.76 V$ , $\frac{2.303 R T}{F} = 0.059 V$ ]

For the redox reaction $Z n (s) + C u^{2 +} (0.1 M) ⟶ Z n^{2 +} (1 M) + C u (s)$ taking place in a cell, $E_{ce l l}^{\circ}$ is $1.10 V$ . $E_{ce l l}$ for the cell will be $(2.303 \frac{R T}{F} = 0.0591)$

The cell, $Zn ∣ Zn^{2 +} (1 M) ∣∣ Cu^{2 +} (1 M) ∣ Cu$ ( $E_{cell}^{\circ} = 1.10 V$ ), was allowed to be completely discharged at 298 K. The relative concentration of $Zn^{2 +}$ to $Cu^{2 +}$ $(\frac{[ Zn ^{2 +} ]}{[ Cu ^{2 +} ]})$ is

Given the electrochemical cell: $E_{Cu^{2 +} ∣ Cu}^{\circ} = + 0.34 V$ $E_{Zn^{2 +} ∣ Zn}^{\circ} = - 0.76 V$ Identify the incorrect statement from the options below for the above cell.

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

Given that the standard potentials ( $E^{\circ}$ ) of $C u^{2 +} / C u$ and $C u^{+} / C u$ are $0.34 V$ and $0.522 V$ respectively, the $E^{\circ}$ of $C u^{2 +} / C u^{+}$ is

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