Answer:
+1.03 V
Explanation:
The standard emf of the voltaic cell is the value of the standard potential of it, which is calculated by the standard reduction potential (E°).
The standard reduction potential is the potential needed for the reduction reaction happen, and it's determined by the reaction with the hydrogen cell (which has E° = 0.0V). The half-reactions of reduction of Ni⁺² and Ag⁺, are:
Ni⁺²(aq) + 2e⁻ → Ni(s) E° = -0.23 V
Ag⁺(aq) + e⁻ → Ag(s) E° = +0.80 V
The value is calculated by a spontaneous reaction, in which the cell with the greater E° is reduced (gain electrons), and the other is oxidized (loses electrons). So, Ag⁺ reduces.
emf = E°reduces - E°oxides
emf = 0.80 - (-0.23)
emf = +1.03 V
Answer:
⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣧⣀⣀⣾⣿⣿⣿⣿
⣿⣿⣿⣿⣿⡏⠉⠛⢿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⡿⣿
⣿⣿⣿⣿⣿⣿⠀⠀⠀⠈⠛⢿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⠿⠛⠉⠁⠀⣿
⣿⣿⣿⣿⣿⣿⣧⡀⠀⠀⠀⠀⠙⠿⠿⠿⠻⠿⠿⠟⠿⠛⠉⠀⠀⠀⠀⠀⣸⣿
⣿⣿⣿⣿⣿⣿⣿⣷⣄⠀⡀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⢀⣴⣿⣿
⣿⣿⣿⣿⣿⣿⣿⣿⣿⠏⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠠⣴⣿⣿⣿⣿
⣿⣿⣿⣿⣿⣿⣿⣿⡟⠀⠀⢰⣹⡆⠀⠀⠀⠀⠀⠀⣭⣷⠀⠀⠀⠸⣿⣿⣿⣿
⣿⣿⣿⣿⣿⣿⣿⣿⠃⠀⠀⠈⠉⠀⠀⠤⠄⠀⠀⠀⠉⠁⠀⠀⠀⠀⢿⣿⣿⣿
⣿⣿⣿⣿⣿⣿⣿⣿⢾⣿⣷⠀⠀⠀⠀⡠⠤⢄⠀⠀⠀⠠⣿⣿⣷⠀⢸⣿⣿⣿
⣿⣿⣿⣿⣿⣿⣿⣿⡀⠉⠀⠀⠀⠀⠀⢄⠀⢀⠀⠀⠀⠀⠉⠉⠁⠀⠀⣿⣿⣿
⣿⣿⣿⣿⣿⣿⣿⣿⣧⠀⠀⠀⠀⠀⠀⠀⠈⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⢹⣿⣿
⣿⣿⣿⣿⣿⣿⣿⣿⣿⠃⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⢸⣿⣿
Explanation:
Due to the cation forms a precipitate by adding H₂S in presence of HCl so this cation must be from group II cations
The choices provided for this question are: Ca²⁺, Mn²⁺, Zn²⁺ and Cd²⁺
So the correct answer will be cadmium since it is the only cation from choices that present in group II cations
If you have other choices you can choose any member of group II cations and divalent like Cu²⁺ , Hg²⁺