Question #e06a0

1 Answer
Jun 7, 2017

Answer:

Here's how to do it.

Explanation:

Start with the half-cell reactions.

#"Cd"^"2+""(aq)" + "2e"^"–" → "Cd(s)";color(white)(l) E^@ = "-0.403 V"#

#"Mg"^"2+"("aq)" + "2e"^"–" → "Mg(s)"; E^@ = "-2.37 V"#

Write the cell reaction

To construct the cell reaction, we must reverse the #"Mg"# half-cell.

#color(white)(mmmmmmmmmmmmmmmmmmmmll)E^@"/V"#
Anode: #color(white)(m)"Mg(s)" → "Mg"^"2+"("aq)" + "2e"^"–" ; color(white)(mmll)"+2.37"#
Cathode: #"Cd"^"2+""(aq)" + "2e"^"–" → "Cd(s)"; color(white)(mmml)"-0.403"#
#stackrel(———————————————————)("Mg(s)" + "Cd"^"2+""(aq)" →"Mg"^"2+""(aq)" + "Cd(s)"); "+1.97"#

Write the cell diagram

The general form for a cell diagram is

#"RED|OX||OX|RED"#

The anode (oxidation) reaction goes on the left, and the cathode (reduction) reaction goes on the right.

The two vertical bars in the middle represent the salt bridge between the two half-cells.

A single vertical bar represents an interface (separation) between two phases.

For this cell, we get

#underbrace("Mg(s)|Mg"^"2+""(aq)")_color(red)("oxidation")"||"underbrace("Cd"^"2+""(aq)|Cd(s)")_color(red)("reduction")#
#color(white)(mmmmmmmll)uparrow#
#color(white)(mmmmmmll)stackrelcolor(blue)("salt bridge")("")#

Here's a cell diagram for a different cell.

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