1864 United States presidential election in Iowa

The 1864 United States presidential election in Iowa took place on November 8, 1864, as part of the 1864 United States presidential election. Iowa voters chose eight representatives, or electors, to the Electoral College, who voted for president and vice president.^[2]

1864 United States presidential election in Iowa

← 1860 November 8, 1864 1868 →
Turnout	19.70% of the total population 0.63 pp[1]
Nominee Abraham Lincoln George B. McClellan Party National Union Democratic Home state Illinois New Jersey Running mate Andrew Johnson George H. Pendleton Electoral vote 8 0 Popular vote 88,500 49,525 Percentage 64.12% 35.88%
County Results Lincoln   50–60%   60–70%   70–80%   80–90%   90–100% McClellan   50–60%   60–70%   70–80%   80–90% Tie   ~50%
President before election Abraham Lincoln Republican Elected President Abraham Lincoln National Union

Iowa was won by the National Union candidate Republican incumbent President Abraham Lincoln of Illinois and his running mate former Senator and Military Governor of Tennessee Andrew Johnson. They defeated the Democratic candidate 4th Commanding General of the United States Army George B. McClellan of New Jersey and his running mate Representative George H. Pendleton of Ohio. Lincoln won the state by a margin of 28.24%.^[2]

Results

1864 United States presidential election in Iowa[2]
Party		Candidate	Running mate	Popular vote		Electoral vote
				Count	%	Count	%
	National Union	Abraham Lincoln of Illinois	Andrew Johnson of Tennessee	88,500	64.12%	8	100.00%
	Democratic	George B. McClellan of New Jersey	George H. Pendleton of Ohio	49,525	35.88%	0	0.00%
Total				138,025	100.00%	8	100.00%

See also

• United States presidential elections in Iowa