Categories > electrochemistry > Electrode potential > Given the half-cell reactions (i) $F{{e}^{2+}}(aq)+2{{e}^{-}}\to Fe\,(s);\,\,E=-0.44\,V$ (ii) $2{{H}^{+}}(aq)+\frac{1}{2}{{O}_{2}}(g)+2{{e}^{-}}\to {{H}_{2}}O\,(l);\,\,{{E}^{0}}=+1.23\,V$ ${{E}^{0}}$ for the reaction $Fe\,(s)+2{{H}^{+}}+\frac{1}{2}{{O}_{2}}(g)\to F{{e}^{2+}}(aq)+{{H}_{2}}O\,(l)$ is

electrochemistry

Question: Given the half-cell reactions (i) $F{{e}^{2+}}(aq)+2{{e}^{-}}\to Fe\,(s);\,\,E=-0.44\,V$ (ii) $2{{H}^{+}}(aq)+\frac{1}{2}{{O}_{2}}(g)+2{{e}^{-}}\to {{H}_{2}}O\,(l);\,\,{{E}^{0}}=+1.23\,V$ ${{E}^{0}}$ for the reaction $Fe\,(s)+2{{H}^{+}}+\frac{1}{2}{{O}_{2}}(g)\to F{{e}^{2+}}(aq)+{{H}_{2}}O\,(l)$ is


1) + 1.67 V
2) – 1.67 V
3) – 0.77 V
4) + 0.77 V
Solution: Explanation: No Explanation
Electrode potential Ecell Nernst equation and ECS

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