$E^{\circ} (O X_{2}, H X^{+} / H X_{2} O) = 1.23 V$ for $O X_{2} + 4 H X^{+} + 4 e X^{-} 2 H X_{2} O$ . The reduction potential of this couple at pH 7 and $p_{O X_{2}} = 1 bar$ at 298 K is (take $2.303 RT / F = 0.0591 V$ )

Answer: A

E = 1.23 + \frac{0.0591}{4} lo g (p_{O X_{2}} [H X^{+}]^{4}) = 1.23 - 0.0591 pH

at 1 bar, since

4/4 = 1

. At pH 7,

E = 1.23 - 0.414 = 0.816 V

. Using

0.0591/4

per pH unit (forgetting the fourth power of

[H X^{+}]

) gives 1.127 V.