Consider the following standard electrode potentials.

     \({\text{S}}{{\text{n}}^{2 + }}{\text{(aq)}} + {\text{2}}{{\text{e}}^ - } \rightleftharpoons {\text{Sn(s)}}\)     \({E^\Theta } =  - 0.14{\text{ V}}\)

     \({{\text{H}}^ + }{\text{(aq)}} + {{\text{e}}^ - } \rightleftharpoons \frac{{\text{1}}}{{\text{2}}}{{\text{H}}_{\text{2}}}{\text{(g)}}\)     \({E^\Theta } = 0.00{\text{ V}}\)

     \({\text{F}}{{\text{e}}^{3 + }}{\text{(aq)}} + {{\text{e}}^ - } \rightleftharpoons {\text{F}}{{\text{e}}^{2 + }}{\text{(aq)}}\)     \({E^\Theta } =  + 0.77{\text{ V}}\)

Which species will reduce H\(^ + \)(aq) to H\(_2\) (g) under standard conditions?

A. Fe\(^{2 + }\)(aq)

B. Sn\(^{2 + }\)(aq)

C. Sn(s)

D. Fe\(^{3 + }\)(aq)

C

60.56% gave the correct answer with the others fairly evenly spread over the three distractors.