A galvanic (voltaic) cell consists of an electrode composed of chromium in a 1.0 M chromium(II) ion solution and another electrode composed of gold in a 1.0 M gold(III) ion solution, connected by a salt bridge. Calculate the standard potential for this cell at 25 °C.

answer: 2.24V

Oh, you're really electrified about galvanic cells, aren't you? Well, let me charge up my humor capacitor and give it a shot!

In this situation, we have chromium hanging out with some chromium(II) ions and gold chilling with gold(III) ions. It sounds like they're having a metallic party!

To calculate the standard potential for this cell, we need to know the reduction potentials for both half-reactions. Unfortunately, I don't have that information on hand, so I can't charge ahead with the calculation. But don't worry, my friend! You can find the reduction potentials in a handy-dandy table of standard reduction potentials. Just make sure the table is fully charged before using it!

Remember, laughter is always a great way to lighten the voltage!

To calculate the standard potential for the galvanic cell, we need to use the standard reduction potentials of chromium(II) and gold(III) half-reactions and apply the Nernst equation.

Step 1: Write the half-reactions:
Chromium(II) ion solution:
Cr^2+ (aq) -> Cr(s) + 2e^-

Gold(III) ion solution:
Au^3+ (aq) + 3e^- -> Au(s)

Step 2: Look up the standard reduction potentials:
The standard reduction potential for the chromium(II) half-reaction is -0.74 V.
The standard reduction potential for the gold(III) half-reaction is +1.50 V.

Step 3: Set up the balanced cell equation:
Cr(s) + Au^3+ (aq) -> Cr^2+ (aq) + Au(s)

Step 4: Calculate the standard potential using the Nernst equation:
The Nernst equation relates the standard potential (E°) with the actual cell potential (E) and the reaction quotient (Q). At standard conditions, Q is equal to 1.

E = E° - (0.0592 V/n) * log(Q)

Since the reaction quotient Q is 1, the Nernst equation simplifies to:

E = E°

Therefore, the standard potential for this galvanic cell is +0.76 V (1.50 V - (-0.74 V)).

You should have included the reduction potentials because they vary from text to text.

Look up reduction potential for Cr(II) to Cr.
Look up reduction potential for Au(III) to Au.
Choose the more negative one, reverse the reaction and change the sign. Then add the two E values.

Au3+(aq) + 3e --> Au(s) E=+1.42

Cr3+ (aq) + 3e -->Cr(aq) E= -0.74 V

Since the standard-reduction potential is positive for Au, the reaction should occur at the cathode and the reaction for Cr3 will occur at the anode.

E(cell)=E(cathode)-E(anode)= 1.42-(-0.74)= 2.16V

I haven't done this in a very long time, so I'm not sure if this is entirely correct, and I am not sure about the half-cell reaction for Au, since I do not have any a reference that I own and I picked it off a site of the interent.

Sorry, it's Cr3+ (aq) + e -->Cr2+(aq) E= -0.41 V not Cr3+ (aq) + 3e -->Cr(aq) E= -0.74 V

E(cell)=E(cathode)-E(anode)= 1.42-(-0.41)= 2.83V

1.83V not 2.83V