Question

1) Determine the potential of a platinum electrode (vs. SHE) when placed in the following solutions at 25°C. A mixture of 0.0141 M potassium permanganate (KMnO4) and 0.0489 M manganese chloride (MnCl2) at pH 3.6

2) When an ion-selective electrode for X was immersed in 0.0500
M XCl, the measured potential was 0.0250 V. What is the
concentration of X when the potential is 0.0400 V? Assume that the
electrode follows the Nernst equation, the temperature is at 25°C,
and that the activity coefficient of X^{+} is 1.

Answer #1

MnO4^{-}(aq) + 5e- +
8H^{+} <--------------> Mn^{2+} (aq)+ 4
H2O(liq) E^{0} = 1.507 V

pH = 3.6

[H+] = 10^-3.6 = 2.51 x 10^-4 M

Ecell = Eo - 0.05916 / n log [Mn+2 / (MnO4-) x H+^8]

E_{cell} = 1.507 - (0.0591/5)
log (0.0489/(0.0141 x ((2.51 x 10^-4 )^8))

= 1.16 V

**E potential = 1.16
V**

2)

X+ + e- -------------> X

no.of electrons = 1

E = Eo + 0.05916 / n log [X+]

= Eo + 0.05916 log [X+]

E = 0.0400 V , [X+] = 0.0500 M

0.04 = Eo + 0.05916 log (0.0500)

Eo = 0.11697 V

when E = 0.0250 V

0.0250 = 0.11697 + 0.05916 log [X+]

**[X+] = 0.0279 M**

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