Question

A galvanic cell based on these half reactions is set up under standard conditions where each solution is 1.00 l and each electrode weighs exactly 100.0 g. how much will the cd electrode weigh when the nonstandard potential of the cell is 0.02543 v

Answer 1

When the nonstandard potential of the cell is 0.02543 v the weight of the cathode is 120.575 g weight.

Step-by-step explanation:

The half reactions are as follows;

Ni²⁺ + 2 e⁻ → Ni E₀ = -0.231 V

Pb²⁺ + 2 e⁻ → Pb E₀ = -0.133 V​

n = 2

Here we have the Nernst equation given by

`E_(cell) = E_(cell)^0 - (RT)/(nF) lnQ = E_(cell)^0 - ((0.0591 V)/(n) )log Q`

Where:

`E_(cell)^0` = 0.098 V

We have, when
`E_(cell)` = 0.02543 V

`E_(cell)` -
`E_(cell)^0` = 0.02543 - 0.098 = -0.07257 V =
`- ((0.0591 V)/(2) )log Q`

log Q = 2.456

Q =
`10^(2.456)` = 285.65

Therefore the number of electron will be

285.65 = the ratio of the Ni²⁺/Pb

Therefore if the initial concentration is 1×10⁻¹ M, we have

`(0.1 + x)/(0.1 - x) = 285.65`

x = 9.93 × 10⁻² Moles

Therefore since 1 mole of Pb = 207.2 g

9.93 × 10⁻² × 207.2 g = 20.575 g

Mass of Pb = 120.575 g.