1888 United States presidential election in Iowa
The 1888 United States presidential election in Iowa took place on November 6, 1888, as part of the 1888 United States presidential election. Voters chose 13 representatives, or electors to the Electoral College, who voted for president and vice president.
 | ||||||||||||||||||||||||||
| ||||||||||||||||||||||||||
| ||||||||||||||||||||||||||
| ||||||||||||||||||||||||||
| Elections in Iowa | ||||||||||
|---|---|---|---|---|---|---|---|---|---|---|
 | ||||||||||
|
||||||||||
|
||||||||||
|
|
||||||||||
Iowa voted for the Republican nominee, Benjamin Harrison, over the Democratic nominee, incumbent President Grover Cleveland. Harrison won the state by a margin of 7.85%.
Results
| 1888 United States presidential election in Iowa[1] | ||||||||
|---|---|---|---|---|---|---|---|---|
| Party | Candidate | Running mate | Popular vote | Electoral vote | ||||
| Count | % | Count | % | |||||
| Republican | Benjamin Harrison of Indiana | Levi Parsons Morton of New York | 211,603 | 52.36% | 13 | 100.00% | ||
| Democratic | Grover Cleveland of New York | Allen Granberry Thurman of Ohio | 179,877 | 44.51% | 0 | 0.00% | ||
| Labor | Alson Streeter of Illinois | Charles E. Cunningham of Arkansas | 9,105 | 2.25% | 0 | 0.00% | ||
| Prohibition | Clinton Bowen Fisk of New Jersey | John Anderson Brooks of Missouri | 3,550 | 0.88% | 0 | 0.00% | ||
| Total | 404,135 | 100.00% | 13 | 100.00% | ||||
Notes
References
- "1888 Presidential General Election Results - Iowa". U.S. Election Atlas. Retrieved 23 December 2013.
State and district results of the 1888 United States presidential election | ||
|---|---|---|
|  | |
This article is issued from Wikipedia. The text is licensed under Creative Commons - Attribution - Sharealike. Additional terms may apply for the media files.