The hydrogen electrode is dipped in a solution of pH=3 at 25^° C. The potential of the electrode will be _________ x 10^-2 V. (frac2.303 RT F=0.059 V)

2 H_( aq .)^++2 e^- -> H_2( g) \ E_ cell = E_ cell ^0-(0.059 / 2) log frac P_ H_2[ H^+]^2 \ =0-0.059 x 3=-0.177 volts. =-17.7 x 10^-2 V \

Home
Test SeriesNEW!
New AppNEW!
About Us

Courses

Virat Batch 2027
Champion Batch 2028
Divine JEE 1-on-1

Contact Us
Knowledge Check

Test Series

The hydrogen electrode is dipped in a solution of pH3 a... | (Chemistry) | DivineJEE

Home
Test SeriesNEW!
New AppNEW!
About Us

Courses

Virat Batch 2027
Champion Batch 2028
Divine JEE 1-on-1

Contact Us
Knowledge Check

Test Series

Chemistry

Exam

Chemistry

The hydrogen electrode is dipped in a solution of $\mathrm{pH}=3$ at $25^{\circ} \mathrm{C}$ . The potential of the electrode will be _________ $\times 10^{-2} \mathrm{~V}$ .

$\left(\frac{2.303 \mathrm{RT}}{\mathrm{F}}=0.059 \mathrm{~V}\right)$

Prev Next

Explanation

$\begin{aligned} & 2 \mathrm{H}_{(\mathrm{aq} .)}^{+}+2 \mathrm{e}^{-} \rightarrow \mathrm{H}_2(\mathrm{~g}) \\ & \mathrm{E}_{\text {cell }}=\mathrm{E}_{\text {cell }}^0-\frac{0.059}{2} \log \frac{\mathrm{P}_{\mathrm{H}_2}}{\left[\mathrm{H}^{+}\right]^2} \\ & =0-0.059 \times 3=-0.177 \text { volts. } \\&=-17.7 \times 10^{-2} \mathrm{~V} \end{aligned}$