A voltaic cell consists of an Mn/Mn2 half-cell and a Pb/Pb2 half-cell. Calculate [Pb2 ] when [Mn2 ] is 1.1 M and E cell is 0.44 V.

Question

A voltaic cell consists of an Mn/Mn2 half-cell and a Pb/Pb2 half-cell. Calculate [Pb2 ] when [Mn2 ] is 1.1 M and E cell is 0.44 V.

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5 months 2021-08-10T01:54:26+00:00 1 Answers 0 views 0

[Pb²⁺] = 2.31×10⁻²¹ M

Explanation:

Let’s write the semi reaction for each half cell:

Pb²⁺ + 2e⁻ ———–> Pb(s)     E° = -0.13 V

Mn²⁺ + 2e⁻ ———-> Mn(s)     E° = -1.18 V

As we can see, the E° of Pb is higher than the E° of the Mn, thus, Pb is reducting and Mn is oxidizing:

Pb²⁺ + 2e⁻ ———–> Pb(s)     E°₁ = -0.13 V

Mn(s) ———> Mn²⁺ + 2e⁻      E°₂ = +1.18 V

E° = E°₁ + E°₂

E° = -0.13 + 1.18 = 1.05 V

Now, we can use the Nerst equation which is:

E = E° – 0.059/n log([Mn²⁺] / [[Pb²⁺])

From here, we just need to replace and then, solve for the [Pb²⁺]:

0.44 = 1.05 – 0.059/2 log(1.1 / x)

0.44 – 1.05 = -0.0295 log(1.1 / x)

-0.61 / -0.0295 = log(1.1 / x)

antlog(20.678) = 1.1 /x

x = [Pb²⁺] = 1.1 / 4.76×10²⁰

[Pb²⁺] = 2.31×10⁻²¹ M

Hope this helps