Question

5.
A solution of concentration CuSo is electrolysed
for 965 s with a current of 20 A and 50% efficiency.
The amount of Cu deposited at cathode will be
[NCERT Pg. 86]
(1) 4.175 g
(2) 3.175 g
(3) 5.35 g
(4) 6.35 g​

Answer 1

Answer:

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Answer 2

The amount of Cu deposited at the cathode will be 3.175g (Option 2)

Given

Solution concentration is made of CuSo₄

Current passed through the solution during electrolysis = 20A

Duration of current passed = 965s

Efficiency = 50%

To Find

Mass in grams of Cu deposited at the cathode

Solution

Formula to find the mass of element deposited in grams = $\frac{M}{nF}$×$it$×$e$ where,

$M$ is the Atomic mass of the element. For $Cu$ it is 63.5g

$n$ is the number of electrons for the anodic or cathodic reaction. Cathodic reduction for $Cu$ is $Cu^{2+}$ + $2e^{-}$ -> $Cu$(s) = 2 electrons

$F$ is the Faraday constant = 96500

$i$ is the current passed in Amperes = 20A

$t$ is the duration of current passed in seconds = 965

$e$ is the electrolytic efficiency = 50% = 0.5

Therefore mass of Cu deposited (g) = (63.5) ÷ (2×96500) × 20 × 965 × 0.5 = 3.175g

Hence the amount of Cu deposited at the cathode will be 3.175g