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

November 8, 1864
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%
President before election Abraham Lincoln Republican Elected President Abraham Lincoln Republican

Iowa was won by the incumbent President Abraham Lincoln (R-Illinois), running with former Senator and Military Governor of Tennessee Andrew Johnson, with 64.12% of the popular vote, against the 4th Commanding General of the United States Army George B. McClellan (D–Pennsylvania), running with Representative George H. Pendleton, with 35.88% of the vote.[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 Pennsylvania	George H. Pendleton of Ohio	49,525	35.88%	0	0.00%
Total				138,025	100.00%	8	100.00%

References

1. "1880 Presidential Election Results Iowa Total Population Turnout".
2. "1864 Presidential Election Results Iowa".