Question

For the reaction 2Cr(s) + 3Pb²⁺ (aq) => 3Pb(s) + 2 Cr³⁺ (aq), what is the value of n in the Nernst equation?

Answer 1

In the reaction 2Cr(s) + 3Pb²⁺ (aq) ⇒ 3Pb(s) + 2 Cr³⁺ (aq), the value of n in the Nernst equation is 6, which represents the number of electrons transferred in the balanced redox reaction.

Step-by-step explanation:

For the reaction 2Cr(s) + 3Pb²⁺ (aq) ⇒ 3Pb(s) + 2 Cr³⁺ (aq), the value of n in the Nernst equation represents the number of moles of electrons transferred in the redox reaction. To determine n, we need to look at the oxidation states of the species involved before and after the reaction.

Anode (oxidation): Cr(s) ⇒ Cr³⁺ (aq) + 3e⁻
Cathode (reduction): Pb²⁺ (aq) + 2e⁻ ⇒ Pb(s)

To balance the electrons, we multiply the cathode half-reaction by 3 and the anode half-reaction by 2, thus getting 6 electrons transferred in both directions:

2Cr(s) + 6e⁻ ⇒ 2Cr³⁺ (aq)
3Pb²⁺ (aq) + 6e⁻ ⇒ 3Pb(s)

Therefore, the value of n for this reaction in the Nernst equation is 6.

Answer 2

The value of n in the Nernst equation for the reaction is 6.

In the Nernst equation, the value of n represents the number of moles of electrons transferred in the balanced chemical equation for the redox reaction.

For the given reaction:

`\[ 2\text{Cr}(s) + 3\text{Pb}^(2+)(\text{aq}) \rightarrow 3\text{Pb}(s) + 2\text{Cr}^(3+)(\text{aq}) \]`

We can see that 6 moles of electrons are involved in the reaction, as indicated by the coefficients in the balanced equation. Therefore, n = 6.

The Nernst equation is given by:

`\[ E = E^\circ - (0.0592)/(n) \log\left(\frac{[\text{Cr}^(3+)]}{[\text{Pb}^(2+)]^3}\right) \]`

Here, n is indeed the number of moles of electrons, which is 6 in this case.