Consider the following E° values : E°_{Fe^{3+}/Fe^{2+}} = +0.77 V, E°_{Sn^{2+}/Sn} = -0.14 V. Under

Q: Why don't we multiply the $E^{\circ}$ of iron by 2 since there are $2Fe^{3+}$ in the equation?

Standard electrode potential ($E^{\circ}$) is an intensive property, meaning it depends only on the nature of the substance and not on the amount. Therefore, stoichiometric coefficients do not affect the $E^{\circ}$ value.

Q: How do I know which one is the cathode and which is the anode?

Look at the overall reaction. Tin ($Sn$) goes from an oxidation state of 0 to +2, meaning it loses electrons (oxidation), so it's the anode. Iron ($Fe^{3+}$) goes from +3 to +2, gaining electrons (reduction), so it's the cathode.

Q: What if the question gave the oxidation potential of tin instead?

If the oxidation potential was given ($+0.14$ V), you could use the formula $E^{\circ}_{cell} = E^{\circ}_{oxidation} + E^{\circ}_{reduction}$. The result would be exactly the same: $0.14 + 0.77 = 0.91$ V.

Q: What does a positive $E^{\circ}_{cell}$ indicate?

A positive standard cell potential indicates that the overall redox reaction is spontaneous under standard conditions, meaning it can generate an electric current.

Q: Can $E^{\circ}_{cell}$ ever be negative for a galvanic cell?

No, a galvanic (or voltaic) cell is designed to produce electrical energy from a spontaneous reaction, so its $E^{\circ}_{cell}$ must be positive. If you calculate a negative value, the reaction is non-spontaneous in the forward direction.

Question

Consider the following $E^{\circ}$ values :
$E^{\circ}_{Fe^{3+}/Fe^{2+}} = +0.77	ext{ V}$
$E^{\circ}_{Sn^{2+}/Sn} = -0.14	ext{ V}$
Under standard conditions, the potential for the reaction
$Sn(s) + 2Fe^{3+}(aq) ightarrow 2Fe^{2+}(aq) + Sn^{2+}(aq)$ is

Options:

(A) 1.68 V

(B) 1.40 V

(C) 0.91 V

(D) 0.63 V

Sigma Percentile · Accepted Answer

The calculation of standard cell potential is one of the most fundamental and rewarding concepts in electrochemistry. It allows us to predict whether a chemical reaction will occur spontaneously and how much electrical driving force it can generate.

In this problem, we are tasked with finding the standard potential for a redox reaction involving tin and iron. Let's break down the thought process step-by-step.

### Analyzing the Setup

Imagine we have a complete electrochemical cell setup right in front of us. On the left side, we have the tin anode. Here, oxidation is taking place as solid tin turns into tin(II) ions.

$$Sn(s) ightarrow Sn^{2+}(aq) + 2e^-$$

On the right side, we have the platinum cathode. In this half-cell, iron(III) ions are gaining electrons and getting reduced to iron(II) ions.

$$2Fe^{3+}(aq) + 2e^- ightarrow 2Fe^{2+}(aq)$$

This entire process is a spontaneous redox reaction that generates an electric current. By identifying the oxidation and reduction half-reactions, we can clearly see which species acts as the anode and which acts as the cathode.

### The Master Equation

To find the standard potential of this entire cell, we need to use a very powerful yet simple formula.

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

This formula essentially tells us the driving force pushing the electrons between the two electrodes. We just need to make absolutely sure that both values we plug in are strictly **standard reduction potentials**.

Look closely at the data provided in the question. We are given the standard reduction potentials for both half-cells. For the iron(III) to iron(II) couple, the value is $+0.77	ext{ V}$. For the tin(II) to solid tin couple, the value is $-0.14	ext{ V}$.

Since iron has the higher reduction potential, it naturally acts as the cathode, forcing tin to act as the anode.

### Final Calculation

Let's substitute the values and get the answer. In our master equation, we will place $0.77$ for the cathode. Then we put down our minus sign, and substitute $-0.14$ for the anode.

$$E^{\circ}_{cell} = 0.77 - (-0.14)$$

Don't make a silly mistake here; we have two negative signs right next to each other, which will mathematically turn into a positive.

$$E^{\circ}_{cell} = 0.77 + 0.14$$

$$E^{\circ}_{cell} = 0.91	ext{ V}$$

So finally, when we add them together, we get **$0.91	ext{ V}$**. This is our final answer.

Also, take a moment to notice that this value is positive. A **positive standard cell potential** indicates that this reaction is completely spontaneous under standard conditions.

### The Intensive Nature of Potentials

There is a catch here, and it's a favorite concept for examiners. Often, students look at the stoichiometry of the reaction, see the coefficient of $2$ in front of iron, and mistakenly multiply the potential by two.

Do not ever do that! **Electrode potential is an intensive property**.

This means it does not depend on the amount of substance or the number of moles reacting. Whether one mole reacts or ten moles react, the potential remains exactly the same. Always trust the raw $E^{\circ}$ values!

Select Answer:

Visualized Solution

Cell Setup

Master Equation

Standard Potentials

Substitution

Final Answer

Intensive Property

The Sigma Insight: Electrochemical Cells

Analyzing the Setup

The Master Equation

Final Calculation

The Intensive Nature of Potentials

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