Question

Use tabulated standard electrode potentials to calculate the standard cell potential for this reaction occurring in an electrochemical cell at 25 ∘C.

Cr(s)+NO−3(aq)4H+(aq)→Cr3+(aq)+NO(g)+2H2O(l)

Express your answer using two decimal places and include the appropriate units.

Answer 1

the first we need to do is find the standar potencial involved in the reaction,

$Cr^{+3} \quad + \quad 3\,e^- \rightleftharpoons Cr \,(s) \quad \quad E^0 = -0.74\,V$

$NO_3^{-} \quad + \quad 4\,H^+ \quad + \quad 3\,e^- \rightleftharpoons NO \,(g) \quad + \quad 2\,H_2O \quad \quad E^0 = 0,96\,V$

now, if you see the reaction, the chrome is oxidized , so we need invert the semi-reaction (notice that it will change the sing of the potential), in order to obtain the original reaction we sum the two semi-reactions and the potential of the cell will be the sum of the potential of the semi-reactions

$Cr \,(s) \rightleftharpoons Cr^{+3} \quad + \quad 3\,e^- \quad \quad E^0 = 0.74\,V$

$NO_3^{-} \quad + \quad 4\,H^+ \quad + \quad 3\,e^- \rightleftharpoons NO \,(g) \quad + \quad 2\,H_2O \quad \quad E^0 = 0,96\,V$

$\mathbf{Cr^{+3} \quad + \quad NO_3^{-} \quad + \quad 4\,H^+\rightleftharpoons Cr\,(s) \quad + \quad NO \,(g) \quad + \quad 2\,H_2O}$

$E_{cell} = E_{red} + E_{oxid} = 0,96\,V \quad + \quad 0,74\,V = \mathbf{1,70\,V}$