Given: {Co}^{3+}+ {e}^{-1} \\to{Co}^{2+} ; {E}^{\0}=+1.81 {V} {Pb}^{4+}+2 {e}^{-} \\to{Pb}^{2+} ; {E}^{\0}=+1.67 {V} {Ce}^{4+}+ {e}^{-} \to{Ce}^{3+} ; {E}=+1.61 {V} {Bi}^{3+}+3 {e}^{-} \\to{Bi} ; {E}^{\0}=+0.20 {V} [12 Apr 2019 Shift 1]

Correct Answer is : Bi3+< Ce4+< Pb4+< Co3+

{Bi}^{3+}< {Ce}^{4+}< {Pb}^{4+}< {Co}^{3+}

{Co}^{3+}< {Ce}^{4+}< {Bi}^{3+}< {Pb}^{4+}

{Ce}^{4+}< {Pb}^{4+}< {Bi}^{3+} <{Co}^{3+}

{Co}^{3+}< {Pb}^{4+}< {Ce}^{4+}< {Bi}^{3+}

Electrochemistry Part 1

Section: Chemistry

Question : 37 of 100

Marks: +1, -0

Given:
Co3++ e−1→Co2+; E0=+1.81V
Pb4++2 e−→Pb2+;E0=+1.67 V
Ce4++ e−→Ce3+;E=+1.61V
Bi3++3 e−→Bi; E0=+0.20 V

[12 Apr 2019 Shift 1]

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