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Chemistry Question on Electrochemistry

Electrode potential of hydrogen electrode is 18 mV,18\, mV, then [H+][H^+] is

A

0.2

B

1

C

2

D

5

Answer

2

Explanation

Solution

Given : EH+/H2=18×10−3 V, [H+]=?E_{_{H^{+}/H_{2}}} = 18\times10^{-3}\,V,\, \left[H^{+}\right]=? Applying Nernst equation, EH+/H2=EH+/H2∘−0.05911log1[H+]E_{_{H^{+}/H_{2}}}=E^{\circ}_{_{H^{+}/H_{2}}}-\frac{0.0591}{1}log \frac{1}{\left[H^{+}\right]} 18×10−3 V=0+0.0591 log[H+]18 \times 10^{-3} \,V = 0 + 0.0591\, log \left[H^{+}\right] 18×10−3 V=0.0591 log[H+]18\times 10^{-3}\, V = 0.0591\, log \left[H^{+}\right] log [H+]=0.3046\left[H^{+}\right] = 0.3046 ∴[H+]=\therefore \left[H^{+}\right]= antilog (0.3046)=2.02=2.0\left(0.3046\right) = 2.02 = 2.0