Write the oxidation and reduction half-reactions for a silver-chromium voltaic cell. (Use the lowest possible coefficients. Include states-of-matter at 1 atm and 25°C in your answer.)
You add the states. You don't say acid or base solution.
Ag ==> Ag^+ + e
3e + CrO4^2 + 8H+- ==> Cr^3+ + 4H2O
To determine the oxidation and reduction half-reactions for a silver-chromium voltaic cell, we first need to identify the oxidized and reduced species. In this case, silver (Ag) will be oxidized to form silver ions (Ag+), while chromium (Cr) will be reduced to form chromium ions (Cr3+).
The oxidation half-reaction involves the loss of electrons, while the reduction half-reaction involves the gain of electrons. We can start by writing the half-reaction for the oxidation of Ag to Ag+:
Ag → Ag+ + e-
Next, let's write the half-reaction for the reduction of Cr to Cr3+:
Cr → Cr3+ + 3e-
Now, let's balance the half-reactions by adding the appropriate coefficients. We want to ensure that the number of electrons gained in the reduction half-reaction is equal to the number of electrons lost in the oxidation half-reaction. To do this, we need to multiply the half-reactions by appropriate coefficients so that the number of electrons balances:
2Ag → 2Ag+ + 2e- (oxidation half-reaction, multiplied by 2)
3Cr → 3Cr3+ + 9e- (reduction half-reaction, multiplied by 3)
Finally, we can write the overall balanced equation for the silver-chromium voltaic cell by adding the half-reactions together:
2Ag + 3Cr → 2Ag+ + 3Cr3+
Note: The states-of-matter at 1 atm and 25°C are usually represented as (s) for solid, (l) for liquid, (g) for gas, and (aq) for aqueous (dissolved in water). However, since the question does not specify the states of the elements, we will leave them as they are.