Question

For the galvanic cell represented by the shorthand notation:

Cr _(s) | Cr²⁺_(aq) || MnO₄^2-_(aq) | H⁺_(aq) | Mn²⁺_(aq) | Pt _(s)

_{** Looking for step by step solution – not as concerned about the answer as much as method**}

(a)   Write the balanced oxidation half reaction

(b)    Write the balanced reduction half reaction

(c)    Write the balanced overall reaction

(d)   Determine E ° _ox

(e)   Determine E ° _Red

(f)    Determine E ° _cell

(g)     What is the value of E _cell if:      

          [Cr²⁺] = [Mn²⁺] = [MnO₄^2-] = 0.10 M,     [H+] = 1.0 x 10 ^-7 M,   and T = 298 K  

(h) What is the pH if:    E_cell = 1.00V,   [Cr²⁺] = [MnO₄^2-] = 0.10 M,   and T = 298 K

Answer 1

a) oxidation reaction is

Cr (s) -------> Cr⁺² +2e-

b) reduction reaction is

MnO4^-   +8H+ +5e- --------> Mn⁺² + 4H2O

c) overall reaction is

5Cr + 2MnO4- +16H+ -------> 5Cr+2 + 2Mn+2 + 8H2O

d) E^{0 f}or oxidation reaction = +0.41V

e) ) E^{0 f}or oxidation reaction = +1.51 V

f) E⁰ cell = E ox + e red

= 0.41 +1.51

= 1.92V

g) According to Nernst equation

E cell = E⁰ cell - (0.0591/n) log [products]/[reactants]

=E⁰ cell - (0.0591/10) log ([Cr+2]⁵[Mn+2]² ) / ([MnO4-]² [H+]¹⁶ )

= 1.91 V - 0.0059 log [0.1]⁷ / 0.1² (1.0x10^-7)¹⁶

= 1.2787 V

g) to calculate pH we need to calculate the [H+]

substituing the given values in Nernst equation

1.00 = 1.91 - (0.059/10) log (0.1)⁷/ (0.1)²[H+]¹⁶

solving for [H+] , we get [H+] = 1.116x10^-10 M