Question

Calculate emf of the cell at 25°C.

Zn_(s) | Zn²⁺ (0.08 M) || Cr³⁺ (0.1 M) | Cr_(s)

`E_(Zn)^0` = −0.76 V,  `E_(Cr)^0` = −0.74 V

Answer 1

Given: `E_(Zn)^0` = −0.76 V,  `E_(Cr)^0` = −0.74 V

To find: Emf of the cell `(E_(cell))`

Formulae: 

1) `E_(cell)^0 = E_(cathode)^0 - E_(anode)^0`

2) `E_(cell) = E_(cell)^0 - (0.0592  V)/n  log_10  [["Product"]]/[["Reactant"]]`

Calculation: 

\[\ce{[Zn_{(s)} -> Zn^{2+}_{(0.08 M)} + 2e-] × 3}\] (oxidation at anode)

\[\ce{[Cr^{3+}_{(0.1 M)} + 3e- -> Cr_{(s)}] × 2}\] (reduction at cathode)

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\[\ce{3Zn_{(s)} + 2Cr^{3+}_{(0.1 M)} -> 3Zn^{2+}_{(0.08 M)} + Cr_{(s)}}\] (overall reaction)

Using formula (1),

`E_(cell)^0 = E_(cathode)^0 - E_(anode)^0`

`E_(cell) = E_(Cr)^0 - E_(Zn)^0`

= −0.74 V − (−0.76 V)

= 0.02 V

Using formula (2),

The cell potential is given by

`E_(cell) = E_(cell)^0 - (0.0592  V)/n log_10  [["Product"]]/[["Reactant"]]`

= `0.02 - (0.0592  V)/6 log_10  (0.08)^3/(0.1)^2`

= `0.02 - 9.867 × 10^-3 log_10  (5.12 × 10^-4)/(1 × 10^-2)`

= `0.02 - 9.867 × 10^-3 × bar 2. 7093`

= 0.02 − 9.867 × 10⁻³ × (−1.2907)

= 0.02 + 0.01273

= 0.03273 V

The emf of the cell is 0.03273 V.