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 $F=96487Cmo{l}^{-1}$

• A. $21.227Jmo{l}^{-1}$
• B. $21.227mJmo{l}^{-1}$
• C. $21.227kJmo{l}^{-1}$
• D. $21.227mo{l}^{-1}$

Answer 1

The correct option is C $21.227kJmo{l}^{-1}$

Let us use the Gibb's equation.
$\Delta G=-nFE$

Here we know $F=96,487Cmo{l}^{-1}$
But we need n and E

Let us quickly write down the half reactions for a Daniel cell
Oxidation: $Zn\to Z{n}^{2+}+2e^{-}$
Reduction: $C{u}^{2+}+2e^{-}\to Cu$

So, n = 2 and we know that $E^0=1.1V$ for a Daniel Cell. (Refer to previous videos if you want to find out how we got this value. )
So, $\Delta G=-2\times 96487\times 1.1=21227Jmo{l}^{-1}=21.227kJmo{l}^{-1}$