First slide

Nernst Equation

Question

The reduction potential of hydrogen half-cell will be negative if

Easy

A

$p (H_{2}) = 1 atm and [H^{+}] = 2.0 M$
B

$p (H_{2}) = 1 atm and [H^{+}] = 1.0 M$
C

$p (H_{2}) = 2 atm and [H^{+}] = 1.0 M$
D

$p (H_{2}) = 2 atm and [H^{+}] = 2.0 M$

Solution

Reduction hydrogen half-cell is $H^{+} |(zM)| Pt (H_{2}); Pressure {pH}_{2}$
Half-cell reaction is $2 H^{+} (aq) + 2 e^{-} \to H_{2} (g)$
Reaction quotient $= Q = {pH}_{2} / {[H^{+}]}^{2} and n = 2$
$E_{cell}^{o} = E_{red}^{o} - \frac{0.0591}{n} \log Q = 0 - \frac{0.0591}{2} \log Q$
${PH}_{2}$ $[H^{+}]$ Q $E_{red}$
1 atm 2.0 M 0.25 +ve
1 atm 1.0 M 1.0 0
2 atm 1.0 M 2.0 -ve
2 atm 2.0 M 0.50 +ve
$E_{red}^{o} = 0.00 V$ for standard hydrogen electrode If $Q > 1, then E_{red}^{o} = - ve$
Thus, correct answer is (c ).

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