1864 United States presidential election in Iowa

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1864 United States presidential election in Iowa

← 1860 November 8, 1864 1868 →
Turnout19.70% of the total population Increase 0.63 pp[1]
  Abraham Lincoln November 1863.jpg GeorgeMcClellan2.jpg
Nominee Abraham Lincoln George B. McClellan
Party National Union Democratic
Home state Illinois Pennsylvania
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

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 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 (DPennsylvania), running with Representative George H. Pendleton, with 35.88% of the vote.[2]

Results[edit]

United States presidential election in Iowa, 1864[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[edit]

  1. ^ "1880 Presidential Election Results Iowa Total Population Turnout".
  2. ^ a b c "1864 Presidential Election Results Iowa".