Chemistry

Chemistry

Problem

The electrode with reaction : Cr2O72−(aq.)+14H+(aq.)+6e⟶2Cr3+(aq.)+7H2O\text{Cr}_2\text{O}_7^{2-}(\text{aq.}) + 14\text{H}^+(\text{aq.}) + 6e \longrightarrow 2\text{Cr}^{3+}(\text{aq.}) + 7\text{H}_2\text{O} ; can be represented as : a. Pt∣H+(aq.),Cr2O72−(aq.)\text{Pt} | \text{H}^+(\text{aq.}), \text{Cr}_2\text{O}_7^{2-}(\text{aq.}) b. Pt∣H+(aq.),Cr2O72−(aq.),Cr3+(aq.)\text{Pt} | \text{H}^+(\text{aq.}), \text{Cr}_2\text{O}_7^{2-}(\text{aq.}), \text{Cr}^{3+}(\text{aq.}) c. PtH2∣H+(aq.),Cr2O72−(aq.)\text{Pt}_{\text{H}_2} | \text{H}^+(\text{aq.}), \text{Cr}_2\text{O}_7^{2-}(\text{aq.}) d. PtH2∣H+(aq.),Cr2O72−(aq.),Cr3+(aq.)\text{Pt}_{\text{H}_2} | \text{H}^+(\text{aq.}), \text{Cr}_2\text{O}_7^{2-}(\text{aq.}), \text{Cr}^{3+}(\text{aq.})

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