Question

The standard cell potential of the following galvanic cell is 1.562 V at 298 K. Zn(s) | Zn2+(aq) || Ag+(aq) | Ag(s) What is the cell potential of the following galvanic cell at 298 K? Zn(s) | Zn2+(aq, 1.00 × 10–3 M) || Ag+(aq, 0.150 M) | Ag(s)

Answer 1

Answer:

E = 1.602 V

Explanation:

Let's consider the following galvanic cell.

Zn(s) | Zn²⁺(aq, 1.00 × 10⁻³ M) || Ag⁺(aq, 0.150 M) | Ag(s)

The corresponding half-reactions are:

Zn(s) → Zn²⁺(aq, 1.00 × 10⁻³ M) + 2 e⁻

2 Ag⁺(aq, 0.150 M) + 2 e⁻ → 2 Ag(s)

The overall reaction is:

Zn(s) + 2 Ag⁺(aq, 0.150 M) → Zn²⁺(aq, 1.00 × 10⁻³ M) + 2 Ag(s)

We can find the cell potential (E) using the Nernst equation.

E = E° - (0.05916/n) . log Q

where,

E°: standard cell potential

n: moles of electrons transferred

Q: reaction quotient

E = E° - (0.05916/n) . log [Zn²⁺]/[Ag⁺]²

E = 1.562 V - (0.05916/2) . log (1.00 × 10⁻³)/(0.150)²

E = 1.602 V

Answer 2

Answer:

E = 1.602v

Explanation:

Use the Nernst Equation => E(non-std) = E⁰(std) – (0.0592/n)logQc …

             Zn⁰(s) => Zn⁺²(aq) + 2 eˉ

2Ag⁺(aq) + 2eˉ=> 2Ag⁰(s)          

_____________________________

Zn⁰(s) + 2Ag⁺(aq) => Zn⁺²(aq) + 2Ag(s)

Given E⁰ = 1.562v

Qc = [Zn⁺²(aq)]/[Ag⁺]² = (1 x 10ˉ³)/(0.150)² = 0.044

E = E⁰ -(0.0592/n)logQc = 1.562v – (0.0592/2)log(0.044) = 1.602v