balance the net ionic equation for the overall reaction.
Pt | Fe2+(aq), Fe3+(aq) || I-(aq) | AgI(s) | Ag(s)
a) Fe3+(aq) + Ag(s) + I-(aq) → Fe2+(aq) + AgI(s)
b) Fe2+(aq) + Ag(s) + I-(aq) → Fe3+(aq) + AgI(s)
c) Fe3+(aq) + AgI(s) → Fe2+(aq) + Ag(s) + I-(aq)
d) Fe2+(aq) + AgI(s) → Fe3+(aq) + Ag(s) + I-(aq)
e) Fe3+(aq) + 2 I-(aq) → Fe2+(aq) + I2(s)
i think it's e, can someone help?
no
To balance the net ionic equation, let's break down the reaction into half-reactions and then combine them:
Half-reaction at the anode (oxidation half-reaction):
Pt → Pt2+ + 2e-
Half-reaction at the cathode (reduction half-reaction):
2Ag+ + 2e- → 2Ag
Combining the half-reactions, we need to make the number of electrons equal on both sides. Multiply the oxidation half-reaction by 2, and the reduction half-reaction by 1:
2Pt → 2Pt2+ + 4e-
2Ag+ + 2e- → 2Ag
Now, we can cancel out the electrons:
2Pt → 2Pt2+ + 4e-
2Ag+ + 4e- → 2Ag
The final net ionic equation is:
2Pt + 2Ag+ → 2Pt2+ + 2Ag
Based on the given answer choices, the correct option is not e. It is option c.
Therefore, the balanced net ionic equation for the overall reaction is:
Fe3+(aq) + AgI(s) → Fe2+(aq) + Ag(s) + I-(aq)
So, the correct answer is option c.