Sapling Week 7/8 question 15

[Pratika Nagpal]

Can someone help me with the steps to solve this.

[Sana Nagori 2H]

I think you use Nernst equation. So you need to find the Enaught and then plug everything in to the equation. For Q you would do .761 on top and then .01 in the denominator.

[Nathan Lao 2I]

This problem involves the Nernst equation. The first step is to find the standard reduction potential of this cell. Then, using the given concentrations, we can solve for Q and plug everything else(such as temp, R, F, etc) into the equation. It''s important to note that the variable n in the question is based on moles of electrons transferred in the equation we get from the standard reduction potential table too.

[Joseph Hsing 2C]

Use the Nernst equation to find the cell potential. E = E naught - RT/nF lnQ. Separate the two half-reactions with the reduction half at the cathode and oxidation half at the anode.

Look up standard reduction potential values and plug into E= E cathode - E anode. (No need to change given signs here)

2 moles of electrons are transferred per mole of reaction when we balanced the half reactions so n=2.

Find Q, convert temp to Kelvins, and lastly plug in all the calculated values and constants into the Nernst equation to find E.

I hope this helped!

[Sara_Lim_2C]

I solved this by using this version of the Nernst Equation: E= (E naught) - (0.05916V/n)log(Q). Use the standard reductions table to find E naught of the half reactions, and use those to find the E naught the whole reaction. Then use the two concentrations to find Q, which is [Cr²⁺]/[Fe²⁺]. n is the number of moles of electrons transferred. We can see in the reaction that 2 moles of electrons are transferred. So then you plug in all those numbers to find E. Hope that helps!

[Natalie 3k]

First you have to find E°cell using the table (Ecathode-Eanode), and with that value you have to use the Nernst equation and plug in the values and that should give you the answer

[Pratika Nagpal]

All of these responses are super helpful. Thank you !