The cell potential for the given cell at 298 K Pt H 2 g , 1 bar H + aq ‖ Cu 2 + aq Cu s is 0 . 31 V . The pH of the acidic solution is found to be 3 , whereas the concentration of Cu 2 + is 10 x M . T

Question

The cell potential for the given cell at $298 K Pt |H_{2} (g, 1 bar)| |H^{+} (aq) ‖ {Cu}^{2 +} (aq)| Cu (s)$ is $0.31 V$ . The $pH$ of the acidic solution is found to be $3$ , whereas the concentration of ${Cu}^{2 +}$ is $10^{x} M$ . The value of $x$ is _________.

(Given: $E_{{Cu}^{2 +} / Cu}^{Θ} = 0.34 V$ and $\frac{2.303 RT}{F} = 0.06 V$ )

TestHub · Accepted Answer

H 2 g + Cu 2 + aq . → 2 H + aq . + Cu s Applying nernst equation E cell = E cell 0 - 0 . 06 n log products Reactants E cell = Cell potential = 0 . 31 V E cell 0 = 0 . 34 p H = 3 , Hence , H + = 10 - 3 0 . 31 = 0 . 34 - 0 . 06 2 log H + 2 Cu 2 + Cu 2 + = 10 - 7 M x = 7

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