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
