1864 United States presidential election in Iowa
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| Turnout | 19.70% of the total population  0.63 pp[1] | |||||||||||||||||||||||||
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 County Results
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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]
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] | ||||||||
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| 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
References
- ^ "1880 Presidential Election Results Iowa Total Population Turnout".
- ^ a b c "1864 Presidential Election Results Iowa".
State results of the 1864 U.S. presidential election | ||
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