Question

Calculate the voltage of the cell Sn_(s) / Sn²⁺(0.02 M) // Ag⁺ (0.01 M) / Ag_(s) at 25 °C.

Given: ${\displaystyle {\text{E}}_{\text{Sn}}^{\circ }}$ = - 0.136, ${\displaystyle {\text{E}}_{\text{Ag}}^{\circ }}$ = 0.800 V

Answer 1

Given: ${\displaystyle {\text{E}}_{\text{Sn}}^{\circ }}$ = - 0.136, ${\displaystyle {\text{E}}_{\text{Ag}}^{\circ }}$ = 0.800 V

To find: Voltage of the cell (E_cell)

Formulae: 

1. ${\displaystyle {\text{E}}_{\text{cell}}^{\circ }={\text{E}}_{\text{cathode}}^{\circ }-{\text{E}}_{\text{anode}}^{\circ }}$
2. ${\displaystyle {\text{E}}_{\text{cell}}={\text{E}}_{\text{cell}}^{\circ }-\frac{0.0592 \text{V}}{\text{n}}{\log }_{10} \frac{\left[\text{Product}\right]}{\left[\text{Reactant}\right]}}$

Calculation: First we write the cell reaction.

$$\mathrm{Sn}{\phantom{A}}_{(\mathrm{s})}⟶\mathrm{Sn}{\phantom{A}}^{2+}(0.02\mathrm{M})+2\mathrm{e}{\phantom{A}}^{-}$$	(oxidation at anode)
[Ag+ (0.01 M) + e– → Ag(s)] × 2	(reduction at cathode)
Overall reaction: $$\mathrm{Sn}{\phantom{A}}_{(\mathrm{s})}+2\mathrm{Ag}{\phantom{A}}^{+}(0.01\mathrm{M})⟶\mathrm{Sn}{\phantom{A}}^{2+}(0.02\mathrm{M})+2\mathrm{Ag}{\phantom{A}}_{(\mathrm{s})}$$

Using formula (1),

${\displaystyle {\text{E}}_{\text{cell}}^{\circ }={\text{E}}_{\text{Ag}}^{\circ }-{\text{E}}_{\text{Sn}}^{\circ }}$ = 0.800 V - (- 0.136 V) = 0.936 V

Using formula (2),

The cell potential is given by

${\displaystyle {\text{E}}_{\text{cell}}={\text{E}}_{\text{cell}}^{\circ }-\frac{0.0592 \text{V}}{\text{n}}{\log }_{10} \frac{\left[{\text{Sn}}^{2+}\right]}{{\left[{\text{Ag}}^{+}\right]}^2}}$

∴ ${\displaystyle {\text{E}}_{\text{cell}}=0.936 \text{V}-\frac{0.0592 \text{V}}{2}{\log }_{10} \frac{0.02}{{\left(0.01\right)}^2}}$

${\displaystyle =0.936 \text{V}-\frac{0.0592 \text{V}}{2}{\log }_{10} 200}$

${\displaystyle =0.936 \text{V}-\frac{0.0592 \text{V}}{2}\times 2.301}$

Calculation using log table:

0.0592 × 2.303

= Antilog₁₀ [log₁₀ 0.0592 + log₁₀ 2.303]

= Antilog₁₀ ${\displaystyle [\overline{2}.7723+0.3623]}$

= Antilog₁₀ ${\displaystyle \left[\overline{1}.1346\right]}$

= 0.1363

= 0.936 V – ${\displaystyle \frac{0.1363 \text{V}}{2}}$  (Using log table)

= 0.936 V – 0.0681 V

= 0.8679 V

The voltage of cell is 0.8679 V.