NEET UG 2019 — Chemistry Physical Chemistry

2019mcqmedium

For a cell involving one electron ${E}_{cell}^{\circleddash }=0.59 V$ at $298 K$ , the equilibrium constant for the cell reaction is:

$[Given that \frac{2.303 RT}{F}=0.059 V at T=298 K]$

Official previous-year question

Held on 30 Apr 2019 · Verified 9 Jul 2026.

Options

A
$1.0\times {10}^{2}$
B
$1.0\times {10}^{5}$
C
$1.0\times {10}^{10}$
D
$1.0\times {10}^{30}$

Did you get this right?

Solution

${E}^{o}=0.59 V$

$\Delta {G}^{o}=-nF{E}^{o};$ Since $(n=1)$

$\Delta {G}^{o}=-RT ln K$

$-1\times F\times 0.59=-RT\times 2.303 log K$

$0.59=\frac{2.303 RT}{F}log K$

$0.59=0.059 log K$

$10=log K$

$K={10}^{10}$

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NEET UG 2019 — Chemistry Physical Chemistry

2019mcqmedium

For a cell involving one electron ${E}_{cell}^{\circleddash }=0.59 V$ at $298 K$ , the equilibrium constant for the cell reaction is:

$[Given that \frac{2.303 RT}{F}=0.059 V at T=298 K]$

Official previous-year question

Held on 30 Apr 2019 · Verified 9 Jul 2026.

Options

A
$1.0\times {10}^{2}$
B
$1.0\times {10}^{5}$
C
$1.0\times {10}^{10}$
D
$1.0\times {10}^{30}$

Did you get this right?

Solution

${E}^{o}=0.59 V$

$\Delta {G}^{o}=-nF{E}^{o};$ Since $(n=1)$

$\Delta {G}^{o}=-RT ln K$

$-1\times F\times 0.59=-RT\times 2.303 log K$

$0.59=\frac{2.303 RT}{F}log K$

$0.59=0.059 log K$

$10=log K$

$K={10}^{10}$

Your note

NEET UG Physical Chemistry in other years

Keep practicing Physical Chemistry

Work through every NEET UG Physical Chemistry PYQ, year by year.

Browse all Chemistry topics

NEET UG 2019 — Chemistry Physical Chemistry

Options

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Solution

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NEET UG Physical Chemistry in other years

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NEET UG 2019 — Chemistry Physical Chemistry

Options

Did you get this right?

Solution

Your note

NEET UG Physical Chemistry in other years

More NEET UG Physical Chemistry questions

Other Chemistry topics for NEET UG

Organic Chemistry

Inorganic Chemistry

Keep practicing Physical Chemistry