Consider the following description of a voltaic cell: A 1.0 M solution of Ni(NO3)2 is placed in a beaker with a strip of nickel metal. A 1.0 M solution of SnSO4 is placed in a second beaker with a strip of Sn. The two beakers are connected byba salt bridge and the two electrodes are linked by a wire to a voltmeter.

a. what is the voltage generated by the cell under standard conditions?

So I looked up the E* of each and found that Nickel was the oxidizer at -0.26, since it is the oxidizer I switched the sign. I found that Sn was 0.15. I added them together and got 0.41 but it said it was wrong. Any help will be appreciated.

I figured it out, the numbers a really 0.236-0.141, I was looking at the wrong Reduction half-reactions.

Thanks for letting me know before I had to look up those numbers.

To determine the voltage generated by the cell under standard conditions, you need to use the standard reduction potentials (E°) of the half-reactions involved in the cell.

In this case, the half-reactions are:
1. At the anode (oxidation): Ni(s) → Ni2+(aq) + 2e-
2. At the cathode (reduction): Sn2+(aq) + 2e- → Sn(s)

You correctly identified the standard reduction potential for the Ni2+(aq)/Ni(s) half-reaction as -0.26 V. However, the standard reduction potential for the Sn2+(aq)/Sn(s) half-reaction is not 0.15 V. The correct value for this half-reaction is -0.14 V.

To calculate the overall cell potential, you need to subtract the reduction potential of the anode (since it is being oxidized) from the reduction potential of the cathode (since it is being reduced):

E°cell = E°cathode - E°anode
= (-0.14 V) - (-0.26 V)
= 0.12 V

Therefore, the voltage generated by the cell under standard conditions is 0.12 V.

a. Which electrode serves as the anode, and which as the cathode? Why?

b. Which electrode gains mass and which loses mass as the cell reaction proceeds?
c. Write the equation for the overall cell reaction.
d. What is the emf generated by the cell under standard conditions?