Question

Eop for the reaction

Mno4- +8H+5e-- Mn2+ +4H2O​

Answer 1

Answer:

E=$E^{0} -\frac{0.0591}{n} Log K_{c}$

For based on above relation $K_{c} =\frac{[Mn^{2+}] }{[Mno_{4}^{-} ][H^{+}]^{8}   }$

And n=5

substitute these values in nernst equation we get

E=$E^{0} -0.01182 Log \frac{[Mn^{2+}] }{[Mno_{4}^{-} ][H^{+} ]^{8}  }$