Question

What is the emf of the cell? $Sn|Sn^{2+} (1\text{M}) || Ag^+ (1\text{M}) | Ag$ [Given $E^0_{Sn^{2+}/Sn} = -0.14\text{V}$ and $E^0_{Ag^+/Ag} = +0.80\text{V}$]

• A. $0.66\text{ V}$
• B. $0.80\text{ V}$
• C. $1.08\text{ V}$
• D. $0.94\text{ V}$

Hint:

Always remember the phrase "Red Cat" (Reduction at Cathode) and that the cathode is the more positive terminal in a galvanic cell. Standard emf must always be positive for a spontaneous reaction.

Answer 1

The Correct Option is D

To calculate the electromotive force (emf) of a standard galvanic cell, we identify the cathode and anode based on their standard reduction potentials ($E^0$). 1. Identifying Cathode and Anode: The electrode with the higher reduction potential acts as the cathode (reduction occurs here), and the one with the lower reduction potential acts as the anode (oxidation occurs here).

• $E^0_{Ag^+/Ag} = +0.80\text{V}$ (Higher value, therefore Cathode) [cite: 74]

• $E^0_{Sn^{2+}/Sn} = -0.14\text{V}$ (Lower value, therefore Anode) [cite: 74]

2. Standard EMF Formula: The standard emf of a cell ($E^0_{cell}$) is calculated using the formula: $$E^0_{cell} = E^0_{cathode} - E^0_{anode}$$

3. Calculation: Substituting the given values: $$E^0_{cell} = (+0.80\text{V}) - (-0.14\text{V})$$ [cite: 74] $$E^0_{cell} = 0.80 + 0.14 = 0.94\text{V}$$ [cite: 74] The calculated emf for the cell is $0.94\text{V}$[cite: 74].