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
Turnout19.70% 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 (DPennsylvania), 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


This article is issued from Wikipedia. The text is licensed under Creative Commons - Attribution - Sharealike. Additional terms may apply for the media files.