Question

Electra is almost bored now. Stupid Nernst making her solve so many problems.

But wait, this is her last one. If she completes this, she is through as an apprentice!Quick. Tell her what is the Gibb's energy of a Daniel Cell.

Use F=96487 C mol−1

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Solution

The correct option is **C** 21.227 kJ mol−1

Let us use the Gibb's equation.

ΔG=−nFE

Here we know F=96,487 C mol−1

But we need n and E

Let us quickly write down the half reactions for a Daniel cell

Oxidation: Zn→Zn2++2e−

Reduction: Cu2++2e−→Cu

So, n = 2 and we know that E0=1.1V for a Daniel Cell. (Refer to previous videos if you want to find out how we got this value. )

So, ΔG=−2×96487×1.1=21227 J mol−1=21.227 kJ mol−1

Let us use the Gibb's equation.

ΔG=−nFE

Here we know F=96,487 C mol−1

But we need n and E

Let us quickly write down the half reactions for a Daniel cell

Oxidation: Zn→Zn2++2e−

Reduction: Cu2++2e−→Cu

So, n = 2 and we know that E0=1.1V for a Daniel Cell. (Refer to previous videos if you want to find out how we got this value. )

So, ΔG=−2×96487×1.1=21227 J mol−1=21.227 kJ mol−1

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