Answer:
[Pb²⁺] = 2.31x10⁻²¹ 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 / 4.76x10²⁰
<h2>
[Pb²⁺] = 2.31x10⁻²¹ M</h2>
Hope this helps
