KVPY SX SB Exam 2018 Question Paper

Question : 101 of 120

Marks: +1, -0

For the electrochemical cell shown below

\Pt|H_2 (P = 1 \atm)|H^{+}(\aq., x M)||Cu^{2+} (\aq., 1.0 M)|Cu(s)

The potential is 0.49 V at 298 K. The pH of the solution is closest to
[Given : Standard reduction potential, E° for Cu

^{2+}

/Cu is 0.34 V]
Gas constant, R is 8.31 J K

^{–1}

mol

^{–1}

Faraday constant, F is 9.65 × 10

^{4}

J V

^{–1}

mol

^{–1}

]

Solution:

anode :

H_2 → 2H^{+} + 2e^{– }

Cathode :

{Cu^{2+} +2e^{-} →Cu(s)}/{H_2+ Cu ^{2+}→Cu+ 2H ^{+} }

E = Eº_\cell - 0.0591/n\log Q⇒0.49 = 0.34 -0.06/2\log[x]^2/1

0.15 = 0.06 \pH

\pH = 2.5

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