Answer:
0.68 V
Explanation:
For anode;
3Mg(s) ---->3Mg^2+(aq) + 6e
For cathode;
2Al^3+(aq) + 6e -----> 2Al(s)
Overall balanced reaction equation;
3Mg(s) + 2Al^3+(aq) ----> 3Mg^2+(aq) + 2Al(s)
Since
E°anode = -2.356 V
E°cathode = -1.676 V
E°cell=-1.676 -(-2.356)
E°cell= 0.68 V
#4 and #5:
To find pH given concentration of H+ or H30+
pH = - log (H+ or H30+ M)
To find pH given concentration of OH-
Since you already found the pH for this (in #4), you subtract #4's answer from 14.
14 - (pH) = pOH
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The chemicals are reactive with one another