Question

What is the voltage of the voltaic cell $Zn|Z{n}^{2+}||C{u}^{2+}|Cu$ at $298K$ if $\left[Z{n}^{2+}\right]=0.2M$ and $\left[C{u}^{2+}\right]=4.0M$?
$C{u}^{2+}+2e^{-}\to Cu{E}^{\circ }=+0.34V$
$Z{n}^{2+}+2e^{-}\to Zn{E}^{\circ }=-0.76V$
[Note: $E=E^{\circ }-\left(0.0591/n\right)\left(\log Q\right)$

• A. $+1.10V$
• B. $-1.10V$
• C. $+1.07V$
• D. $+1.14V$
• E. $-1.07V$

Answer 1

The correct option is D $+1.14V$

The representation of the voltaic cell shows that $Zn$ is oxidised to $Z{n}^{2+}$, whereas, $C{u}^{2+}$ is reduced to $Cu$.

These two phenomenons can be written by two equations-

$C{u}^{2+}$$+$$2e^{-}$$\to$$Cu$..... $E^o$$=$ $+0.34V$, (since $C{u}^{2+}$ is reduced)..... $eq\left(1\right)$

and

$Z{n}^{2+}$$+$$2e^{-}$$\to$$Zn$..... $E^o$$=$ $-0.76V$

But the voltaic cell represents that $Zn$ is oxidised to $Z{n}^{2+}$, hence the equation must be written as-

$Zn$$\to$$Z{n}^{2+}$$+$$2e^{-}$..... $E^o$$=$ $+0.76V$, (since $Zn$ is oxidised)......$eq\left(2\right)$

Adding $eq\left(1\right)$ and $eq\left(2\right)$, we get,

$C{u}^{2+}$$+$$Zn$$+$$2e^{-}$$\to$$Cu$$+$$Z{n}^{2+}$$+$$2e^{-}$...... $E^o$$=$$0.34+0.76$$=$$1.10V$

Or, $C{u}^{2+}$$+$$Zn$$\to$$Cu$$+$$Z{n}^{2+}$..... $E^o$$=$ $+1.10V$

We also know the formula-

$E$ $=$ $E^o$ $-$ $\frac{0.0591}{n}$$\left(logQ\right)$

Q can be written as $Q$$=$$\frac{\left[Z{n}^{2+}\right]}{\left[C{u}^{2+}\right]}$ and $n$ is the number of electrons, that is $2$

Putting the values of all the known quantities in the given equation we get,

$E$ $=$ $1.10$ $-$ $\frac{0.0591}{2}$$log$$\frac{\left[0.2\right]}{\left[4.0\right]}$

Thus, $E$$=$$1.10-(-0.038)$

$E$$=$$1.138V$, that is $1.14V$

Option D is the correct answer.