Practice Exercise 19.07 What is the overall cell reaction of a galvanic cell emp

Question

Practice Exercise 19.07 What is the overall cell reaction of a galvanic cell employing the following half-reactions? NiO2(s) + 2H2O + 2e" z Ni(OH)2(S) + 2OH(aq), E°Nio2= 0.49 V; Fe(OH)2(S) + 2e z Fe(s) + 2OH-(aq), E° Fe(OH)2= - 0.88 v. O NiO2(S) + Fe(s) + 2H20 - Ni(OH)2(aq) + Fe(OH)2(aq) O Ni(OH)2(s) + Fe(OH)2(S) — NiO2(S) + Fe(s) + 2H20 O NiO2(S) + Fe(s) + 2H2O Ni(OH)2(S) + Fe(OH)2(s) O NiO2(s) + 2Fe(s) + 2H20 Ni(OH)2(s) + 2Fe(OH)2(s) What is the standard cell potential of a galvanic cell? E° cell = v Eºcell = The number of significant digits is set to 3; the tolerance is +/-1 in the 3rd significant digit

Explanation / Answer

we know that

oxidation takes place at anode

reduction takes place at cathode

now

the half reaction with more positive reduction potential is cathode

so

cathode :

NiO2 (s) + 2 H20 + 2e ---> Ni(OH)2 + 2 OH- Eo = 0.49 V

anode : oxidation

Fe (s) + 2 OH- ---> Fe (OH)2 + 2e- Eo = -0.88 V

so the net reaction is

NiO2 (s) + Fe (s) + 2 H2O ---> Ni(OH)2 (s) + Fe(OH)2 (s)

2)

now

Eo cell = Eo cathode - Eo anode

Eo cell = 0.49 - (-0.88)

Eo cell = 1.37 V

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