6N 13

$E_{cell} = E_{cell}^{\circ}-\frac{RT}{nF}\ln Q$

rachelle1K
Posts: 109
Joined: Sat Sep 07, 2019 12:16 am

6N 13

6N.13 Calculate the reaction quotient, Q, for the following cell reactions, given the measured values of the cell potential. Balance the chemical equations by using the smallest whole-number coefficients. (a) Pt(s) Sn4+(aq),Sn2+(aq) Pb4+(aq),Pb2+(aq) C(gr), Ecell = +1.33 V.
I'm having a little trouble solving this problem. I know that we use the Nerst Equation, and therefore we must find E standard.

For the standard reduction potential, I got 1.67 for the cathode, and .15 for the anode. Therefore, the E standard is 1.52.

Next, I did 1.33-1.52= -.19

=(-.19(2))/(-.05916)=logQ
= 6.42 = logQ
$10^{6.42}=Q$

However, the answer key says Q= 10 ^6

What am I doing wrong?

Kylie Lim 4G
Posts: 110
Joined: Sat Aug 17, 2019 12:15 am

Re: 6N 13

The 0.19V should be divided by (RT/NF) which I got to be 0.012839. The result is 14.798 which is approximately 15.

Jason Wu 1E
Posts: 101
Joined: Thu Jul 25, 2019 12:15 am

Re: 6N 13

I got the same result as you... I'm not sure but based on previous posts for the same question they have said there is an error in the solutions manual. However, the answer that they got was also different which confuses me even more :/

Brooke Yasuda 2J
Posts: 102
Joined: Sat Jul 20, 2019 12:17 am

Re: 6N 13

I believe you forgot that the n value is equal to 2, not 1. So you would have t divide .05916 by 2 to get the correct answer

Ruby Richter 2L
Posts: 103
Joined: Thu Jul 25, 2019 12:17 am

Re: 6N 13

I don't understand how lnQ=15 gives you Q=10^6. Could someone explain?

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