For the galvanic cell represented by the shorthand notation: Cr (s ) | Cr 2+ (aq
ID: 490678 • Letter: F
Question
For the galvanic cell represented by the shorthand notation:
Cr (s) | Cr2+(aq) || MnO42-(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+] = [MnO42-] = 0.10 M, [H+] = 1.0 x 10 -7 M, and T = 298 K
(h) What is the pH if: Ecell = 1.00V, [Cr2+] = [MnO42-] = 0.10 M, and T = 298 K
Explanation / Answer
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) E0 for oxidation reaction = +0.41V
e) ) E0 for 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