For the cell Cu(s) | Cu²⁺ (aq)(0.1M) || Ag⁺ (aq)(0.01M) | Ag(s)
The cell potential E₁ = 0.3095V
For the cell Cu(s) | Cu²⁺ (aq)(0.01M) || Ag⁺ (aq)(0.001M) | Ag(s)
The cell potential = _____ *10⁻² V. (Round off to the Nearest Integer).
[Use : (2.303RT/F) = 0.059 ]

9 Views|Posted 5 months ago
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5 months ago

E_cell = E? _cell - (0.059/2)log { [Cu²? ]/ [Ag? ]²}
E? = E? _cell - (0.059/2)log [0.1/ (0.01)²] = 0.3095
E? _cell = 0.3095 + 0.0885 V
E? _cell = 0.398

Again, E? = E? _cell - (0.059/2)log [10? ²/ (10? ³)²]
E? = 0.398 - (0.059/2) x 4 V

E? = 0.28 V
E? = 28 x 10? ² V

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Ecell = 1.05 - (0.059 / 2) log ( [Ni²? ] / [Ag? ]²)
= 1.05 - (0.059 / 2) log ( [10? ³] / [10? ³]²)
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There is a misprint in the question. The E? cell is incorrectly given as 10.5 V. This should have been 1.05 volt.

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