$$Ag^+_{(aq)} + e^- \longrightarrow Ag_{(s)} \hspace{0.5cm} E° = 0.80 V$$
$$Ni_{(s)}^2+_{(aq)} + 2 e^- \longrightarrow Ni_{(s)} \hspace{0.5cm} E° = -0.25 V$$
The overall cell reaction is:
$$Ag^+_{(aq)} + Ni_{(s)} \longrightarrow Ag_{(s)} + Ni_{(s)}^2+_{(aq)}$$
The standard cell potential is:
$$E°_{cell} = E°_{cathode} - E°_{anode}$$
$$E°_{cell} = 0.80 V - (-0.25 V) = 1.05 V$$
Therefore, the voltage of the galvanic cell is 1.05 V.
