Answer:
Anode (oxidation): Cr(s) ⇒ Cr³⁺(aq) + 3 e⁻
Cathode (reduction): Ag⁺(aq) + 1 e⁻ ⇒ Ag(s)
Explanation:
Let's consider the notation of a galvanic cell.
Cr(s) | Cr³⁺(aq) || Ag⁺(aq) | Ag(s)
On the left, it is represented the anode (oxidation) and on the right, it is represented the cathode (reduction).
The half-reactions are:
Anode (oxidation): Cr(s) ⇒ Cr³⁺(aq) + 3 e⁻
Cathode (reduction): Ag⁺(aq) + 1 e⁻ ⇒ Ag(s)
To have the global reaction, we have to multiply the reduction by 3 (so the number of electrons gained and lost are the same) and add both half-reactions.
Global reaction: Cr(s) + 3 Ag⁺(aq) ⇒ Cr³⁺(aq) + 3 Ag(s)
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This type interaction is called the London dispersion force. These are temporary forces that happens when electrons of two adjacent atoms occupy positions that will result to temporary dipoles. The constant motion of the electrons will cause to form an instantaneous dipole when the electrons are unsymmetrically distributed. These forces are present in nonpolar substances.