Calculate the cell potential for the following reaction as written at 79C given that [Zn^+2] = 0.819 M and [Ni^+2] = 0.014 M.

Zn + Ni^+2 = Zn^+2 + Ni

I know: E=E -RT/nF ln Q
reduction potential for Ni= -0.26
reduction potential for Zn = -0.76
so E standard potential = 0.5
E= 0.5 - (8.315)(352)/(2)(96480) ln 0.819/0.014.

I'm getting E= 1.989 and my answer is wrong.

The set up looks ok but I would use 8.314 for R and 96,485 for F. If I run through that I obtain 0.438.

E= 0.438?

I got the same answer (1.987) when I used the numbers you mentioned. Is the E standard potential = -0.26 -(-0.76)?
Am I doing that right?(its cathode-Anode) right?

Ecell = 0.5v. You are correct. I plugged in your numbers and obtained 0.438 with them, also. This is like looking for a needle in haystack. The BEST think to have done was type in your work and let someone look at it. As it is both are floundering. Here is your work.

E= 0.5 - (8.315)(352)/(2)(96480) ln 0.819/0.014.

In pieces:
8.314*352/(2*96480) = 0.01517
0.819/0.014 = 58.5
ln 58.5 = 4.060
0.01517*4.069 = 0.0617 and
0.5-0.0617 = 0.438
Check your work. One of us is punching the wrong keys.

Got it, you are right, it's 0.439. Thank you so much, I was doing an easy math mistake (substracting before multiplying). Thanks again.

To calculate the cell potential, you need to use the Nernst equation:

E = E° - (RT / nF) * ln(Q)

where:
E is the cell potential at the given conditions
E° is the standard cell potential
R is the gas constant (8.315 J/(mol·K))
T is the temperature in Kelvin
n is the number of moles of electrons transferred in the balanced equation
F is Faraday's constant (96480 C/mol)
ln is the natural logarithm
Q is the reaction quotient

In this case, the balanced equation is:

Zn + Ni^+2 → Zn^+2 + Ni

The number of moles of electrons transferred (n) is 2 because one Zn atom loses 2 electrons and one Ni^+2 ion gains 2 electrons.

Now let's calculate the cell potential:

1. Convert the temperature from Celsius to Kelvin: 79°C + 273.15 = 352.15 K.

2. Calculate the standard cell potential (E°):

E° = E°(cathode) - E°(anode)
= 0.00 V - (-0.76 V)
= 0.76 V

3. Calculate the reaction quotient (Q):

Q = [Zn^+2] / [Ni^+2]
= 0.819 M / 0.014 M
= 58.5

4. Plug the values into the Nernst equation:

E = 0.76 V - [(8.315 J/(mol·K)) * (352.15 K)] / (2 * 96480 C/mol) * ln(58.5)

Calculating this expression:

E = 0.76 V - (2931.44 J / 19296 C) * ln(58.5)
E = 0.76 V - 0.15148 V * ln(58.5)
E ≈ 0.76 V - 0.15148 V * 4.068
E ≈ 0.76 V - 0.61694 V
E ≈ 0.14306 V

So, the cell potential for the given reaction at 79°C is approximately 0.14306 V.